{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0001","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"cdbedc6035d66a6cec85077a4999bba8f1ea80e29268fb687c2294d9b657bb6d","family_id":"qf-function-limit-affine","seed":202610082001,"semantic_template":"qf-function-limit-affine/{\"inputs\":[\"value_b\",\"0\"],\"operation\":\"affine_punctured_limit\"}"},"fields":{"title":"Affine Limit on a Punctured Real Domain","stem":"On R∖{0}, f(x)=(0)x+(0). Find its punctured two-sided limit as x→0, without assigning f(0).","answer":{"kind":"short_text","value":"0","explanation":"The affine approach value is m*a+b=0. The original function's punctured domain does not require a defined center value."},"skills":["Affine Limit on a Punctured Real Domain"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0002","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b2262c3960bf9f90af34d3c73f4e08df336c3571f7a2ef8801874c3a9d72d73a","family_id":"qf-function-limit-affine","seed":202610082002,"semantic_template":"qf-function-limit-affine/{\"inputs\":[\"value_b\",\"-11\"],\"operation\":\"affine_punctured_limit\"}"},"fields":{"title":"Affine Limit on a Punctured Real Domain","stem":"On R∖{0}, f(x)=(-2)x+(-11). Find its punctured two-sided limit as x→0, without assigning f(0).","answer":{"kind":"short_text","value":"-11","explanation":"The affine approach value is m*a+b=-11. The original function's punctured domain does not require a defined center value."},"skills":["Affine Limit on a Punctured Real Domain"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0003","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"92fbbe40d661a539ec550ed82b65b2ef9e330ae3a69468f22f3972a735449885","family_id":"qf-function-limit-affine","seed":202610082003,"semantic_template":"qf-function-limit-affine/{\"inputs\":[\"3\",\"-10\",\"3\"],\"operation\":\"affine_punctured_limit\"}"},"fields":{"title":"Affine Limit on a Punctured Real Domain","stem":"On R∖{3}, f(x)=(3)x+(-10). Find its punctured two-sided limit as x→3, without assigning f(3).","answer":{"kind":"short_text","value":"-1","explanation":"The affine approach value is m*a+b=-1. The original function's punctured domain does not require a defined center value."},"skills":["Affine Limit on a Punctured Real Domain"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0004","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b291d5d3d4e4f451dbc5c2da15e5307e5b42fcc93c9ed6426216b78d87a0ae6a","family_id":"qf-function-limit-affine","seed":202610082004,"semantic_template":"qf-function-limit-affine/{\"inputs\":[\"-4\",\"-9\",\"4\"],\"operation\":\"affine_punctured_limit\"}"},"fields":{"title":"Affine Limit on a Punctured Real Domain","stem":"On R∖{4}, f(x)=(-4)x+(-9). Find its punctured two-sided limit as x→4, without assigning f(4).","answer":{"kind":"short_text","value":"-25","explanation":"The affine approach value is m*a+b=-25. The original function's punctured domain does not require a defined center value."},"skills":["Affine Limit on a Punctured Real Domain"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0005","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e9e9a548c798b2fad475bb992e9457e67bd83b57c750c80bc76bf7e916d3edbf","family_id":"qf-function-limit-affine","seed":202610082005,"semantic_template":"qf-function-limit-affine/{\"inputs\":[\"5\",\"-8\",\"1\"],\"operation\":\"affine_punctured_limit\"}"},"fields":{"title":"Affine Limit on a Punctured Real Domain","stem":"On R∖{1}, f(x)=(5)x+(-8). Find its punctured two-sided limit as x→1, without assigning f(1).","answer":{"kind":"short_text","value":"-3","explanation":"The affine approach value is m*a+b=-3. The original function's punctured domain does not require a defined center value."},"skills":["Affine Limit on a Punctured Real Domain"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0006","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6d417ca19d6baa604662910591618c39062c8f5ac482859fbb19c7f758e8cf06","family_id":"qf-function-limit-affine","seed":202610082006,"semantic_template":"qf-function-limit-affine/{\"inputs\":[\"-6\",\"-7\",\"2\"],\"operation\":\"affine_punctured_limit\"}"},"fields":{"title":"Affine Limit on a Punctured Real Domain","stem":"On R∖{2}, f(x)=(-6)x+(-7). Find its punctured two-sided limit as x→2, without assigning f(2).","answer":{"kind":"short_text","value":"-19","explanation":"The affine approach value is m*a+b=-19. The original function's punctured domain does not require a defined center value."},"skills":["Affine Limit on a Punctured Real Domain"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0007","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f43144297a99ab5cf2f08d7cd6f861c896c4947419bd5452b1c30b9bbe2bc1f7","family_id":"qf-function-limit-affine","seed":202610082007,"semantic_template":"qf-function-limit-affine/{\"inputs\":[\"7\",\"-6\",\"3\"],\"operation\":\"affine_punctured_limit\"}"},"fields":{"title":"Affine Limit on a Punctured Real Domain","stem":"On R∖{3}, f(x)=(7)x+(-6). Find its punctured two-sided limit as x→3, without assigning f(3).","answer":{"kind":"short_text","value":"15","explanation":"The affine approach value is m*a+b=15. The original function's punctured domain does not require a defined center value."},"skills":["Affine Limit on a Punctured Real Domain"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0008","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b41a57e6aae4d0b6afe26454568760f6376962823f904087c4fbcf032f6fe487","family_id":"qf-function-limit-affine","seed":202610082008,"semantic_template":"qf-function-limit-affine/{\"inputs\":[\"-8\",\"-5\",\"4\"],\"operation\":\"affine_punctured_limit\"}"},"fields":{"title":"Affine Limit on a Punctured Real Domain","stem":"On R∖{4}, f(x)=(-8)x+(-5). Find its punctured two-sided limit as x→4, without assigning f(4).","answer":{"kind":"short_text","value":"-37","explanation":"The affine approach value is m*a+b=-37. The original function's punctured domain does not require a defined center value."},"skills":["Affine Limit on a Punctured Real Domain"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0009","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f824a11cd02a4c8adaf46bd5507fe949963e0bb5122568ec33fd58c8d457e119","family_id":"qf-function-limit-affine","seed":202610082009,"semantic_template":"qf-function-limit-affine/{\"inputs\":[\"9\",\"-4\",\"1\"],\"operation\":\"affine_punctured_limit\"}"},"fields":{"title":"Affine Limit on a Punctured Real Domain","stem":"On R∖{1}, f(x)=(9)x+(-4). Find its punctured two-sided limit as x→1, without assigning f(1).","answer":{"kind":"short_text","value":"5","explanation":"The affine approach value is m*a+b=5. The original function's punctured domain does not require a defined center value."},"skills":["Affine Limit on a Punctured Real Domain"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0010","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9e703fbe7c4169b3137224198adf2b54f5e8784d61e6113262cbd6f089453880","family_id":"qf-function-limit-affine","seed":202610082010,"semantic_template":"qf-function-limit-affine/{\"inputs\":[\"-10\",\"-3\",\"2\"],\"operation\":\"affine_punctured_limit\"}"},"fields":{"title":"Affine Limit on a Punctured Real Domain","stem":"On R∖{2}, f(x)=(-10)x+(-3). Find its punctured two-sided limit as x→2, without assigning f(2).","answer":{"kind":"short_text","value":"-23","explanation":"The affine approach value is m*a+b=-23. The original function's punctured domain does not require a defined center value."},"skills":["Affine Limit on a Punctured Real Domain"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0011","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3806430bf1326208ae5aa9456cb3668056412bc5dc626c3293dec70f49ff31a5","family_id":"qf-function-limit-affine","seed":202610082011,"semantic_template":"qf-function-limit-affine/{\"inputs\":[\"11\",\"-2\",\"3\"],\"operation\":\"affine_punctured_limit\"}"},"fields":{"title":"Affine Limit on a Punctured Real Domain","stem":"On R∖{3}, f(x)=(11)x+(-2). Find its punctured two-sided limit as x→3, without assigning f(3).","answer":{"kind":"short_text","value":"31","explanation":"The affine approach value is m*a+b=31. The original function's punctured domain does not require a defined center value."},"skills":["Affine Limit on a Punctured Real Domain"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0012","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b6c647ef5489389c50d76e127ab1927090b3fa8e10190c4af19dabf21cb3898e","family_id":"qf-function-limit-affine","seed":202610082012,"semantic_template":"qf-function-limit-affine/{\"inputs\":[\"-12\",\"-1\",\"4\"],\"operation\":\"affine_punctured_limit\"}"},"fields":{"title":"Affine Limit on a Punctured Real Domain","stem":"On R∖{4}, f(x)=(-12)x+(-1). Find its punctured two-sided limit as x→4, without assigning f(4).","answer":{"kind":"short_text","value":"-49","explanation":"The affine approach value is m*a+b=-49. The original function's punctured domain does not require a defined center value."},"skills":["Affine Limit on a Punctured Real Domain"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0013","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2af3a730f8da34de1589e3ebadff6b969e42ea04d717310a384e1efcf2c6faba","family_id":"qf-function-limit-affine","seed":202610082013,"semantic_template":"qf-function-limit-affine/{\"inputs\":[\"13\",\"0\",\"1\"],\"operation\":\"affine_punctured_limit\"}"},"fields":{"title":"Affine Limit on a Punctured Real Domain","stem":"On R∖{1}, f(x)=(13)x+(0). Find its punctured two-sided limit as x→1, without assigning f(1).","answer":{"kind":"short_text","value":"13","explanation":"The affine approach value is m*a+b=13. The original function's punctured domain does not require a defined center value."},"skills":["Affine Limit on a Punctured Real Domain"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0014","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fc74fae2f833ddd0dbd692fe340939da678e5788373e4b6e71c53b9f35eefc67","family_id":"qf-function-limit-affine","seed":202610082014,"semantic_template":"qf-function-limit-affine/{\"inputs\":[\"-14\",\"1\",\"2\"],\"operation\":\"affine_punctured_limit\"}"},"fields":{"title":"Affine Limit on a Punctured Real Domain","stem":"On R∖{2}, f(x)=(-14)x+(1). Find its punctured two-sided limit as x→2, without assigning f(2).","answer":{"kind":"short_text","value":"-27","explanation":"The affine approach value is m*a+b=-27. The original function's punctured domain does not require a defined center value."},"skills":["Affine Limit on a Punctured Real Domain"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0015","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2d6b177f43c204a5e254e84452d4a67051e87ae8a3d9752803088e836dc05c38","family_id":"qf-function-limit-affine","seed":202610082015,"semantic_template":"qf-function-limit-affine/{\"inputs\":[\"15\",\"2\",\"3\"],\"operation\":\"affine_punctured_limit\"}"},"fields":{"title":"Affine Limit on a Punctured Real Domain","stem":"On R∖{3}, f(x)=(15)x+(2). Find its punctured two-sided limit as x→3, without assigning f(3).","answer":{"kind":"short_text","value":"47","explanation":"The affine approach value is m*a+b=47. The original function's punctured domain does not require a defined center value."},"skills":["Affine Limit on a Punctured Real Domain"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0016","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"de0b1fd6201d6c03987bced69b627bf9c7eeab400e2cb3407221189dbbf8da6d","family_id":"qf-function-limit-affine","seed":202610082016,"semantic_template":"qf-function-limit-affine/{\"inputs\":[\"-16\",\"3\",\"4\"],\"operation\":\"affine_punctured_limit\"}"},"fields":{"title":"Affine Limit on a Punctured Real Domain","stem":"On R∖{4}, f(x)=(-16)x+(3). Find its punctured two-sided limit as x→4, without assigning f(4).","answer":{"kind":"short_text","value":"-61","explanation":"The affine approach value is m*a+b=-61. The original function's punctured domain does not require a defined center value."},"skills":["Affine Limit on a Punctured Real Domain"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0017","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0ada97bdc002219837285a7f440c5a57843a6e4b63d5cdd3febeec72782510e5","family_id":"qf-function-limit-affine","seed":202610082017,"semantic_template":"qf-function-limit-affine/{\"inputs\":[\"17\",\"4\",\"1\"],\"operation\":\"affine_punctured_limit\"}"},"fields":{"title":"Affine Limit on a Punctured Real Domain","stem":"On R∖{1}, f(x)=(17)x+(4). Find its punctured two-sided limit as x→1, without assigning f(1).","answer":{"kind":"short_text","value":"21","explanation":"The affine approach value is m*a+b=21. The original function's punctured domain does not require a defined center value."},"skills":["Affine Limit on a Punctured Real Domain"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0018","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"555cb717f18f9105f2e2e32b82ce73e205445c1f153e608072464d19a156c897","family_id":"qf-function-limit-affine","seed":202610082018,"semantic_template":"qf-function-limit-affine/{\"inputs\":[\"-18\",\"5\",\"2\"],\"operation\":\"affine_punctured_limit\"}"},"fields":{"title":"Affine Limit on a Punctured Real Domain","stem":"On R∖{2}, f(x)=(-18)x+(5). Find its punctured two-sided limit as x→2, without assigning f(2).","answer":{"kind":"short_text","value":"-31","explanation":"The affine approach value is m*a+b=-31. The original function's punctured domain does not require a defined center value."},"skills":["Affine Limit on a Punctured Real Domain"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0019","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1000a825a9412c6a5363a86896e2035e7601b3d4a89d32b6e7ef282d222dfb98","family_id":"qf-function-limit-affine","seed":202610082019,"semantic_template":"qf-function-limit-affine/{\"inputs\":[\"19\",\"6\",\"3\"],\"operation\":\"affine_punctured_limit\"}"},"fields":{"title":"Affine Limit on a Punctured Real Domain","stem":"On R∖{3}, f(x)=(19)x+(6). Find its punctured two-sided limit as x→3, without assigning f(3).","answer":{"kind":"short_text","value":"63","explanation":"The affine approach value is m*a+b=63. The original function's punctured domain does not require a defined center value."},"skills":["Affine Limit on a Punctured Real Domain"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0020","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"dc98d6e22a60c15b4bdd8c693e385e6e44993df2781bda87ce37c975b3a4a607","family_id":"qf-function-limit-affine","seed":202610082020,"semantic_template":"qf-function-limit-affine/{\"inputs\":[\"-20\",\"7\",\"4\"],\"operation\":\"affine_punctured_limit\"}"},"fields":{"title":"Affine Limit on a Punctured Real Domain","stem":"On R∖{4}, f(x)=(-20)x+(7). Find its punctured two-sided limit as x→4, without assigning f(4).","answer":{"kind":"short_text","value":"-73","explanation":"The affine approach value is m*a+b=-73. The original function's punctured domain does not require a defined center value."},"skills":["Affine Limit on a Punctured Real Domain"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0021","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"93baba8f0400980435dfb17a350d1239858a4bec594cfa3ca30103e4a708856e","family_id":"qf-function-limit-affine","seed":202610082021,"semantic_template":"qf-function-limit-affine/{\"inputs\":[\"21\",\"8\",\"1\"],\"operation\":\"affine_punctured_limit\"}"},"fields":{"title":"Affine Limit on a Punctured Real Domain","stem":"On R∖{1}, f(x)=(21)x+(8). Find its punctured two-sided limit as x→1, without assigning f(1).","answer":{"kind":"short_text","value":"29","explanation":"The affine approach value is m*a+b=29. The original function's punctured domain does not require a defined center value."},"skills":["Affine Limit on a Punctured Real Domain"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0022","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0d91179852d593793d7102b349dfa59948d8adf2248d1cd47644226536c3b23e","family_id":"qf-function-limit-affine","seed":202610082022,"semantic_template":"qf-function-limit-affine/{\"inputs\":[\"-22\",\"9\",\"2\"],\"operation\":\"affine_punctured_limit\"}"},"fields":{"title":"Affine Limit on a Punctured Real Domain","stem":"On R∖{2}, f(x)=(-22)x+(9). Find its punctured two-sided limit as x→2, without assigning f(2).","answer":{"kind":"short_text","value":"-35","explanation":"The affine approach value is m*a+b=-35. The original function's punctured domain does not require a defined center value."},"skills":["Affine Limit on a Punctured Real Domain"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0023","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a5d5f0d46129e9171fe04c16000fc0388117893d97dd2dcef5e8247304e7c914","family_id":"qf-function-limit-affine","seed":202610082023,"semantic_template":"qf-function-limit-affine/{\"inputs\":[\"23\",\"10\",\"3\"],\"operation\":\"affine_punctured_limit\"}"},"fields":{"title":"Affine Limit on a Punctured Real Domain","stem":"On R∖{3}, f(x)=(23)x+(10). Find its punctured two-sided limit as x→3, without assigning f(3).","answer":{"kind":"short_text","value":"79","explanation":"The affine approach value is m*a+b=79. The original function's punctured domain does not require a defined center value."},"skills":["Affine Limit on a Punctured Real Domain"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0024","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f303515f21fc82ffac311dde3911160b0f49617765ce986302b44829d0824031","family_id":"qf-function-limit-affine","seed":202610082024,"semantic_template":"qf-function-limit-affine/{\"inputs\":[\"-24\",\"11\",\"4\"],\"operation\":\"affine_punctured_limit\"}"},"fields":{"title":"Affine Limit on a Punctured Real Domain","stem":"On R∖{4}, f(x)=(-24)x+(11). Find its punctured two-sided limit as x→4, without assigning f(4).","answer":{"kind":"short_text","value":"-85","explanation":"The affine approach value is m*a+b=-85. The original function's punctured domain does not require a defined center value."},"skills":["Affine Limit on a Punctured Real Domain"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0025","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"50401eb0e87710da3f980b55ae2af0d99925a1f7afa90efc32dd7a9a83a8c40e","family_id":"qf-function-limit-affine","seed":202610082025,"semantic_template":"qf-function-limit-affine/{\"inputs\":[\"25\",\"12\",\"1\"],\"operation\":\"affine_punctured_limit\"}"},"fields":{"title":"Affine Limit on a Punctured Real Domain","stem":"On R∖{1}, f(x)=(25)x+(12). Find its punctured two-sided limit as x→1, without assigning f(1).","answer":{"kind":"short_text","value":"37","explanation":"The affine approach value is m*a+b=37. The original function's punctured domain does not require a defined center value."},"skills":["Affine Limit on a Punctured Real Domain"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0026","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3b6ba83a67be63711bd55d3218a5b757d3b029c38b577246f01757c056fab905","family_id":"qf-function-limit-removable-quotient","seed":202610082026,"semantic_template":"qf-function-limit-removable-quotient/{\"inputs\":[\"-12\"],\"operation\":\"removable_quotient_limit\"}"},"fields":{"title":"Punctured Limit of a Removable Quotient","stem":"On R∖{-12}, f(x)=(x²−(-12)²)/(x−(-12)). Find its punctured two-sided limit as x→-12.","answer":{"kind":"short_text","value":"-24","explanation":"For x≠a, cancel the nonzero factor x−a to obtain x+a; its approach value is 2a=-24. This does not define the original quotient at the center or perform division by zero."},"skills":["Punctured Limit of a Removable Quotient"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0027","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b5b26abd1c577af8326e49a0f3b251fec9c5e1f33b1e8bf0442ccda9eb6a07d0","family_id":"qf-function-limit-removable-quotient","seed":202610082027,"semantic_template":"qf-function-limit-removable-quotient/{\"inputs\":[\"-11\"],\"operation\":\"removable_quotient_limit\"}"},"fields":{"title":"Punctured Limit of a Removable Quotient","stem":"On R∖{-11}, f(x)=(x²−(-11)²)/(x−(-11)). Find its punctured two-sided limit as x→-11.","answer":{"kind":"short_text","value":"-22","explanation":"For x≠a, cancel the nonzero factor x−a to obtain x+a; its approach value is 2a=-22. This does not define the original quotient at the center or perform division by zero."},"skills":["Punctured Limit of a Removable Quotient"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0028","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d5c2758e72613984bc81ba727b44a309b1ded1d6323fe7a6f92ac33b4d089aa4","family_id":"qf-function-limit-removable-quotient","seed":202610082028,"semantic_template":"qf-function-limit-removable-quotient/{\"inputs\":[\"-10\"],\"operation\":\"removable_quotient_limit\"}"},"fields":{"title":"Punctured Limit of a Removable Quotient","stem":"On R∖{-10}, f(x)=(x²−(-10)²)/(x−(-10)). Find its punctured two-sided limit as x→-10.","answer":{"kind":"short_text","value":"-20","explanation":"For x≠a, cancel the nonzero factor x−a to obtain x+a; its approach value is 2a=-20. This does not define the original quotient at the center or perform division by zero."},"skills":["Punctured Limit of a Removable Quotient"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0029","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7151ffe12ec3d4e061c2178adc6dab0beb5a93752eb21e73942dd9e874c317af","family_id":"qf-function-limit-removable-quotient","seed":202610082029,"semantic_template":"qf-function-limit-removable-quotient/{\"inputs\":[\"-9\"],\"operation\":\"removable_quotient_limit\"}"},"fields":{"title":"Punctured Limit of a Removable Quotient","stem":"On R∖{-9}, f(x)=(x²−(-9)²)/(x−(-9)). Find its punctured two-sided limit as x→-9.","answer":{"kind":"short_text","value":"-18","explanation":"For x≠a, cancel the nonzero factor x−a to obtain x+a; its approach value is 2a=-18. This does not define the original quotient at the center or perform division by zero."},"skills":["Punctured Limit of a Removable Quotient"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0030","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a948efde19e960fbf992ea020a35c6ecf9ec5e6bd2252fd219982eb314351340","family_id":"qf-function-limit-removable-quotient","seed":202610082030,"semantic_template":"qf-function-limit-removable-quotient/{\"inputs\":[\"-8\"],\"operation\":\"removable_quotient_limit\"}"},"fields":{"title":"Punctured Limit of a Removable Quotient","stem":"On R∖{-8}, f(x)=(x²−(-8)²)/(x−(-8)). Find its punctured two-sided limit as x→-8.","answer":{"kind":"short_text","value":"-16","explanation":"For x≠a, cancel the nonzero factor x−a to obtain x+a; its approach value is 2a=-16. This does not define the original quotient at the center or perform division by zero."},"skills":["Punctured Limit of a Removable Quotient"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0031","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f43e3ac8525a6477aa312f4af55c26f46a4bdde2882f05eca2bd7f594d23f543","family_id":"qf-function-limit-removable-quotient","seed":202610082031,"semantic_template":"qf-function-limit-removable-quotient/{\"inputs\":[\"-7\"],\"operation\":\"removable_quotient_limit\"}"},"fields":{"title":"Punctured Limit of a Removable Quotient","stem":"On R∖{-7}, f(x)=(x²−(-7)²)/(x−(-7)). Find its punctured two-sided limit as x→-7.","answer":{"kind":"short_text","value":"-14","explanation":"For x≠a, cancel the nonzero factor x−a to obtain x+a; its approach value is 2a=-14. This does not define the original quotient at the center or perform division by zero."},"skills":["Punctured Limit of a Removable Quotient"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0032","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3cb18e0ddac60432d29a42b119e99506ff6ede7959b05dabaae7ad1f04688887","family_id":"qf-function-limit-removable-quotient","seed":202610082032,"semantic_template":"qf-function-limit-removable-quotient/{\"inputs\":[\"-6\"],\"operation\":\"removable_quotient_limit\"}"},"fields":{"title":"Punctured Limit of a Removable Quotient","stem":"On R∖{-6}, f(x)=(x²−(-6)²)/(x−(-6)). Find its punctured two-sided limit as x→-6.","answer":{"kind":"short_text","value":"-12","explanation":"For x≠a, cancel the nonzero factor x−a to obtain x+a; its approach value is 2a=-12. This does not define the original quotient at the center or perform division by zero."},"skills":["Punctured Limit of a Removable Quotient"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0033","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"718cdf9a035f01f1dd5c0cc1b8f1c5548e75f6df7dd854315032ae356f80232e","family_id":"qf-function-limit-removable-quotient","seed":202610082033,"semantic_template":"qf-function-limit-removable-quotient/{\"inputs\":[\"-5\"],\"operation\":\"removable_quotient_limit\"}"},"fields":{"title":"Punctured Limit of a Removable Quotient","stem":"On R∖{-5}, f(x)=(x²−(-5)²)/(x−(-5)). Find its punctured two-sided limit as x→-5.","answer":{"kind":"short_text","value":"-10","explanation":"For x≠a, cancel the nonzero factor x−a to obtain x+a; its approach value is 2a=-10. This does not define the original quotient at the center or perform division by zero."},"skills":["Punctured Limit of a Removable Quotient"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0034","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"01d6cfb9a0640b66b5d1c42021d73aea71a32d3bfd02cb5981b3de637b0a19e2","family_id":"qf-function-limit-removable-quotient","seed":202610082034,"semantic_template":"qf-function-limit-removable-quotient/{\"inputs\":[\"-4\"],\"operation\":\"removable_quotient_limit\"}"},"fields":{"title":"Punctured Limit of a Removable Quotient","stem":"On R∖{-4}, f(x)=(x²−(-4)²)/(x−(-4)). Find its punctured two-sided limit as x→-4.","answer":{"kind":"short_text","value":"-8","explanation":"For x≠a, cancel the nonzero factor x−a to obtain x+a; its approach value is 2a=-8. This does not define the original quotient at the center or perform division by zero."},"skills":["Punctured Limit of a Removable Quotient"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0035","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"aa9806cdd44b41092db279e2c6c045377cefa3725f5db7699df2982f2b3c9e24","family_id":"qf-function-limit-removable-quotient","seed":202610082035,"semantic_template":"qf-function-limit-removable-quotient/{\"inputs\":[\"-3\"],\"operation\":\"removable_quotient_limit\"}"},"fields":{"title":"Punctured Limit of a Removable Quotient","stem":"On R∖{-3}, f(x)=(x²−(-3)²)/(x−(-3)). Find its punctured two-sided limit as x→-3.","answer":{"kind":"short_text","value":"-6","explanation":"For x≠a, cancel the nonzero factor x−a to obtain x+a; its approach value is 2a=-6. This does not define the original quotient at the center or perform division by zero."},"skills":["Punctured Limit of a Removable Quotient"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0036","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7dacbf115ea0e07cbd6edf17a7a0f2e649ce1a0f48b6b35ee122cd7a30563327","family_id":"qf-function-limit-removable-quotient","seed":202610082036,"semantic_template":"qf-function-limit-removable-quotient/{\"inputs\":[\"-2\"],\"operation\":\"removable_quotient_limit\"}"},"fields":{"title":"Punctured Limit of a Removable Quotient","stem":"On R∖{-2}, f(x)=(x²−(-2)²)/(x−(-2)). Find its punctured two-sided limit as x→-2.","answer":{"kind":"short_text","value":"-4","explanation":"For x≠a, cancel the nonzero factor x−a to obtain x+a; its approach value is 2a=-4. This does not define the original quotient at the center or perform division by zero."},"skills":["Punctured Limit of a Removable Quotient"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0037","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3c189902013b458925568024e65fe57de4bae02821ceac9cb52ec6b88dc7524a","family_id":"qf-function-limit-removable-quotient","seed":202610082037,"semantic_template":"qf-function-limit-removable-quotient/{\"inputs\":[\"-1\"],\"operation\":\"removable_quotient_limit\"}"},"fields":{"title":"Punctured Limit of a Removable Quotient","stem":"On R∖{-1}, f(x)=(x²−(-1)²)/(x−(-1)). Find its punctured two-sided limit as x→-1.","answer":{"kind":"short_text","value":"-2","explanation":"For x≠a, cancel the nonzero factor x−a to obtain x+a; its approach value is 2a=-2. This does not define the original quotient at the center or perform division by zero."},"skills":["Punctured Limit of a Removable Quotient"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0038","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0f0f65c8ad81441cad984940d019ae4174441f41091364511b8c2089788923d9","family_id":"qf-function-limit-removable-quotient","seed":202610082038,"semantic_template":"qf-function-limit-removable-quotient/{\"inputs\":[\"0\"],\"operation\":\"removable_quotient_limit\"}"},"fields":{"title":"Punctured Limit of a Removable Quotient","stem":"On R∖{0}, f(x)=(x²−(0)²)/(x−(0)). Find its punctured two-sided limit as x→0.","answer":{"kind":"short_text","value":"0","explanation":"For x≠a, cancel the nonzero factor x−a to obtain x+a; its approach value is 2a=0. This does not define the original quotient at the center or perform division by zero."},"skills":["Punctured Limit of a Removable Quotient"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0039","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"380e134244c4b7a6cffb1fcefd9f6931ab97460bdb76b8a330ab3cca9f8f2d77","family_id":"qf-function-limit-removable-quotient","seed":202610082039,"semantic_template":"qf-function-limit-removable-quotient/{\"inputs\":[\"1\"],\"operation\":\"removable_quotient_limit\"}"},"fields":{"title":"Punctured Limit of a Removable Quotient","stem":"On R∖{1}, f(x)=(x²−(1)²)/(x−(1)). Find its punctured two-sided limit as x→1.","answer":{"kind":"short_text","value":"2","explanation":"For x≠a, cancel the nonzero factor x−a to obtain x+a; its approach value is 2a=2. This does not define the original quotient at the center or perform division by zero."},"skills":["Punctured Limit of a Removable Quotient"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0040","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a53e21c0cf3e836c767c04a4d63d2718ae443a8e5de568e2b7440d485f4899b5","family_id":"qf-function-limit-removable-quotient","seed":202610082040,"semantic_template":"qf-function-limit-removable-quotient/{\"inputs\":[\"2\"],\"operation\":\"removable_quotient_limit\"}"},"fields":{"title":"Punctured Limit of a Removable Quotient","stem":"On R∖{2}, f(x)=(x²−(2)²)/(x−(2)). Find its punctured two-sided limit as x→2.","answer":{"kind":"short_text","value":"4","explanation":"For x≠a, cancel the nonzero factor x−a to obtain x+a; its approach value is 2a=4. This does not define the original quotient at the center or perform division by zero."},"skills":["Punctured Limit of a Removable Quotient"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0041","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d73208e6d6490860eca14864d1ae9009913134d68efc2b3006a1cade3cb0feef","family_id":"qf-function-limit-removable-quotient","seed":202610082041,"semantic_template":"qf-function-limit-removable-quotient/{\"inputs\":[\"3\"],\"operation\":\"removable_quotient_limit\"}"},"fields":{"title":"Punctured Limit of a Removable Quotient","stem":"On R∖{3}, f(x)=(x²−(3)²)/(x−(3)). Find its punctured two-sided limit as x→3.","answer":{"kind":"short_text","value":"6","explanation":"For x≠a, cancel the nonzero factor x−a to obtain x+a; its approach value is 2a=6. This does not define the original quotient at the center or perform division by zero."},"skills":["Punctured Limit of a Removable Quotient"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0042","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"422ce7ef95efb4d55079f974aa3eb846ee7fe28c3a8cb07c032425ddcd422d81","family_id":"qf-function-limit-removable-quotient","seed":202610082042,"semantic_template":"qf-function-limit-removable-quotient/{\"inputs\":[\"4\"],\"operation\":\"removable_quotient_limit\"}"},"fields":{"title":"Punctured Limit of a Removable Quotient","stem":"On R∖{4}, f(x)=(x²−(4)²)/(x−(4)). Find its punctured two-sided limit as x→4.","answer":{"kind":"short_text","value":"8","explanation":"For x≠a, cancel the nonzero factor x−a to obtain x+a; its approach value is 2a=8. This does not define the original quotient at the center or perform division by zero."},"skills":["Punctured Limit of a Removable Quotient"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0043","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2184001fc0da152587991d54707ee4a46671dff4453a79181737dd259a98a9e8","family_id":"qf-function-limit-removable-quotient","seed":202610082043,"semantic_template":"qf-function-limit-removable-quotient/{\"inputs\":[\"5\"],\"operation\":\"removable_quotient_limit\"}"},"fields":{"title":"Punctured Limit of a Removable Quotient","stem":"On R∖{5}, f(x)=(x²−(5)²)/(x−(5)). Find its punctured two-sided limit as x→5.","answer":{"kind":"short_text","value":"10","explanation":"For x≠a, cancel the nonzero factor x−a to obtain x+a; its approach value is 2a=10. This does not define the original quotient at the center or perform division by zero."},"skills":["Punctured Limit of a Removable Quotient"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0044","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"10a68da19bc07e25ef22d0b2ad262ae726d48aa3c17daa6af02652925c215617","family_id":"qf-function-limit-removable-quotient","seed":202610082044,"semantic_template":"qf-function-limit-removable-quotient/{\"inputs\":[\"6\"],\"operation\":\"removable_quotient_limit\"}"},"fields":{"title":"Punctured Limit of a Removable Quotient","stem":"On R∖{6}, f(x)=(x²−(6)²)/(x−(6)). Find its punctured two-sided limit as x→6.","answer":{"kind":"short_text","value":"12","explanation":"For x≠a, cancel the nonzero factor x−a to obtain x+a; its approach value is 2a=12. This does not define the original quotient at the center or perform division by zero."},"skills":["Punctured Limit of a Removable Quotient"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0045","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7eec95450e59e28afe5f72a7b0369f9da54067704fabc2d6b85a0ac9b070b6c9","family_id":"qf-function-limit-removable-quotient","seed":202610082045,"semantic_template":"qf-function-limit-removable-quotient/{\"inputs\":[\"7\"],\"operation\":\"removable_quotient_limit\"}"},"fields":{"title":"Punctured Limit of a Removable Quotient","stem":"On R∖{7}, f(x)=(x²−(7)²)/(x−(7)). Find its punctured two-sided limit as x→7.","answer":{"kind":"short_text","value":"14","explanation":"For x≠a, cancel the nonzero factor x−a to obtain x+a; its approach value is 2a=14. This does not define the original quotient at the center or perform division by zero."},"skills":["Punctured Limit of a Removable Quotient"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0046","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"81c69e5934bd327f8e860ed7d1d42fc5ea633b30477fa980ff50cd2d346d35c3","family_id":"qf-function-limit-removable-quotient","seed":202610082046,"semantic_template":"qf-function-limit-removable-quotient/{\"inputs\":[\"8\"],\"operation\":\"removable_quotient_limit\"}"},"fields":{"title":"Punctured Limit of a Removable Quotient","stem":"On R∖{8}, f(x)=(x²−(8)²)/(x−(8)). Find its punctured two-sided limit as x→8.","answer":{"kind":"short_text","value":"16","explanation":"For x≠a, cancel the nonzero factor x−a to obtain x+a; its approach value is 2a=16. This does not define the original quotient at the center or perform division by zero."},"skills":["Punctured Limit of a Removable Quotient"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0047","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f81a46de7b74fc8ff0b69be4f243f57476e4912b5df7b755d56c8b26f554e884","family_id":"qf-function-limit-removable-quotient","seed":202610082047,"semantic_template":"qf-function-limit-removable-quotient/{\"inputs\":[\"9\"],\"operation\":\"removable_quotient_limit\"}"},"fields":{"title":"Punctured Limit of a Removable Quotient","stem":"On R∖{9}, f(x)=(x²−(9)²)/(x−(9)). Find its punctured two-sided limit as x→9.","answer":{"kind":"short_text","value":"18","explanation":"For x≠a, cancel the nonzero factor x−a to obtain x+a; its approach value is 2a=18. This does not define the original quotient at the center or perform division by zero."},"skills":["Punctured Limit of a Removable Quotient"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0048","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5ad4dfe3a39f85a00d9910dee2cc2098458e545068a6f2227b527a3c2fa77b7b","family_id":"qf-function-limit-removable-quotient","seed":202610082048,"semantic_template":"qf-function-limit-removable-quotient/{\"inputs\":[\"10\"],\"operation\":\"removable_quotient_limit\"}"},"fields":{"title":"Punctured Limit of a Removable Quotient","stem":"On R∖{10}, f(x)=(x²−(10)²)/(x−(10)). Find its punctured two-sided limit as x→10.","answer":{"kind":"short_text","value":"20","explanation":"For x≠a, cancel the nonzero factor x−a to obtain x+a; its approach value is 2a=20. This does not define the original quotient at the center or perform division by zero."},"skills":["Punctured Limit of a Removable Quotient"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0049","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ee5c1badec0cc5e4f977bf0990fbfbceb85eb12bec003bb1c5b731c6202cd7e3","family_id":"qf-function-limit-removable-quotient","seed":202610082049,"semantic_template":"qf-function-limit-removable-quotient/{\"inputs\":[\"11\"],\"operation\":\"removable_quotient_limit\"}"},"fields":{"title":"Punctured Limit of a Removable Quotient","stem":"On R∖{11}, f(x)=(x²−(11)²)/(x−(11)). Find its punctured two-sided limit as x→11.","answer":{"kind":"short_text","value":"22","explanation":"For x≠a, cancel the nonzero factor x−a to obtain x+a; its approach value is 2a=22. This does not define the original quotient at the center or perform division by zero."},"skills":["Punctured Limit of a Removable Quotient"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0050","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3146abb6ee2e63bbc3197b7994ed7657ec070d0a6da54f01760d873a1a68c934","family_id":"qf-function-limit-removable-quotient","seed":202610082050,"semantic_template":"qf-function-limit-removable-quotient/{\"inputs\":[\"12\"],\"operation\":\"removable_quotient_limit\"}"},"fields":{"title":"Punctured Limit of a Removable Quotient","stem":"On R∖{12}, f(x)=(x²−(12)²)/(x−(12)). Find its punctured two-sided limit as x→12.","answer":{"kind":"short_text","value":"24","explanation":"For x≠a, cancel the nonzero factor x−a to obtain x+a; its approach value is 2a=24. This does not define the original quotient at the center or perform division by zero."},"skills":["Punctured Limit of a Removable Quotient"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0051","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7820112533bc6ebd6aecfd822accb22f3f4e861e16df07518e2b0fbe19409b33","family_id":"qf-function-limit-affine-delta","seed":202610082051,"semantic_template":"qf-function-limit-affine-delta/{\"inputs\":[\"constant_affine\"],\"operation\":\"affine_delta_validity\"}"},"fields":{"title":"Affine δ Witness with Strict Punctured Interior","stem":"For arbitrary real A,L, define f(x)=L+(0)(x−A) for x≠A on R∖{A}. With ε=1/1>0 and δ=5/1>0, does 0<|x−A|<δ guarantee |f(x)−L|<ε?","answer":{"kind":"short_text","value":"Yes","explanation":"Check |m|δ≤ε by exact cross products, giving Yes. Equality is permitted because the punctured input distance is strictly below δ. If m=0 every positive δ works. The target is L, not an undefined f(A)."},"skills":["Affine δ Witness with Strict Punctured Interior"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0052","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3cadf59daa5369b34abb94abec513209fecc14ea8494e5b8683a11eae79576e3","family_id":"qf-function-limit-affine-delta","seed":202610082052,"semantic_template":"qf-function-limit-affine-delta/{\"inputs\":[\"1\",\"1/1\",\"1/1\"],\"operation\":\"affine_delta_validity\"}"},"fields":{"title":"Affine δ Witness with Strict Punctured Interior","stem":"For arbitrary real A,L, define f(x)=L+(-1)(x−A) for x≠A on R∖{A}. With ε=2/2>0 and δ=2/2>0, does 0<|x−A|<δ guarantee |f(x)−L|<ε?","answer":{"kind":"short_text","value":"Yes","explanation":"Check |m|δ≤ε by exact cross products, giving Yes. Equality is permitted because the punctured input distance is strictly below δ. If m=0 every positive δ works. The target is L, not an undefined f(A)."},"skills":["Affine δ Witness with Strict Punctured Interior"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0053","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a7ac81fdc71f12d4af01dea6370c792818087b4016f7a7e387c6df980223f96b","family_id":"qf-function-limit-affine-delta","seed":202610082053,"semantic_template":"qf-function-limit-affine-delta/{\"inputs\":[\"2\",\"1/1\",\"1/4\"],\"operation\":\"affine_delta_validity\"}"},"fields":{"title":"Affine δ Witness with Strict Punctured Interior","stem":"For arbitrary real A,L, define f(x)=L+(2)(x−A) for x≠A on R∖{A}. With ε=3/3>0 and δ=3/12>0, does 0<|x−A|<δ guarantee |f(x)−L|<ε?","answer":{"kind":"short_text","value":"Yes","explanation":"Check |m|δ≤ε by exact cross products, giving Yes. Equality is permitted because the punctured input distance is strictly below δ. If m=0 every positive δ works. The target is L, not an undefined f(A)."},"skills":["Affine δ Witness with Strict Punctured Interior"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0054","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e5a0b8bde43121838298813ef1f60dd105fbd23207e444619dd3ecd87d9346ae","family_id":"qf-function-limit-affine-delta","seed":202610082054,"semantic_template":"qf-function-limit-affine-delta/{\"inputs\":[\"3\",\"1/4\",\"1/6\"],\"operation\":\"affine_delta_validity\"}"},"fields":{"title":"Affine δ Witness with Strict Punctured Interior","stem":"For arbitrary real A,L, define f(x)=L+(-3)(x−A) for x≠A on R∖{A}. With ε=1/4>0 and δ=2/12>0, does 0<|x−A|<δ guarantee |f(x)−L|<ε?","answer":{"kind":"short_text","value":"No","explanation":"Check |m|δ≤ε by exact cross products, giving No. Equality is permitted because the punctured input distance is strictly below δ. If m=0 every positive δ works. The target is L, not an undefined f(A)."},"skills":["Affine δ Witness with Strict Punctured Interior"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0055","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"920da152c57e8ba26aaed2bfd8d183adfd3e5b138da8619d64d1f0e59730cca6","family_id":"qf-function-limit-affine-delta","seed":202610082055,"semantic_template":"qf-function-limit-affine-delta/{\"inputs\":[\"4\",\"2/1\",\"1/2\"],\"operation\":\"affine_delta_validity\"}"},"fields":{"title":"Affine δ Witness with Strict Punctured Interior","stem":"For arbitrary real A,L, define f(x)=L+(4)(x−A) for x≠A on R∖{A}. With ε=2/1>0 and δ=2/4>0, does 0<|x−A|<δ guarantee |f(x)−L|<ε?","answer":{"kind":"short_text","value":"Yes","explanation":"Check |m|δ≤ε by exact cross products, giving Yes. Equality is permitted because the punctured input distance is strictly below δ. If m=0 every positive δ works. The target is L, not an undefined f(A)."},"skills":["Affine δ Witness with Strict Punctured Interior"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0056","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f29411d88e26ac03f9eeab709c5219a0812a9ca3cf91849ef7427d610f0331c1","family_id":"qf-function-limit-affine-delta","seed":202610082056,"semantic_template":"qf-function-limit-affine-delta/{\"inputs\":[\"5\",\"3/2\",\"3/20\"],\"operation\":\"affine_delta_validity\"}"},"fields":{"title":"Affine δ Witness with Strict Punctured Interior","stem":"For arbitrary real A,L, define f(x)=L+(-5)(x−A) for x≠A on R∖{A}. With ε=3/2>0 and δ=3/20>0, does 0<|x−A|<δ guarantee |f(x)−L|<ε?","answer":{"kind":"short_text","value":"Yes","explanation":"Check |m|δ≤ε by exact cross products, giving Yes. Equality is permitted because the punctured input distance is strictly below δ. If m=0 every positive δ works. The target is L, not an undefined f(A)."},"skills":["Affine δ Witness with Strict Punctured Interior"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0057","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"467b1bac7ec821fb553b3f16aff7ee26d54095b3e5732db3fae08e5f889f0e07","family_id":"qf-function-limit-affine-delta","seed":202610082057,"semantic_template":"qf-function-limit-affine-delta/{\"inputs\":[\"6\",\"1/3\",\"1/9\"],\"operation\":\"affine_delta_validity\"}"},"fields":{"title":"Affine δ Witness with Strict Punctured Interior","stem":"For arbitrary real A,L, define f(x)=L+(6)(x−A) for x≠A on R∖{A}. With ε=1/3>0 and δ=2/18>0, does 0<|x−A|<δ guarantee |f(x)−L|<ε?","answer":{"kind":"short_text","value":"No","explanation":"Check |m|δ≤ε by exact cross products, giving No. Equality is permitted because the punctured input distance is strictly below δ. If m=0 every positive δ works. The target is L, not an undefined f(A)."},"skills":["Affine δ Witness with Strict Punctured Interior"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0058","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"89cd88f4b015a15fccc1b4d6bff7ab13f27f4762755a5d13fd1bd00ad90f6c16","family_id":"qf-function-limit-affine-delta","seed":202610082058,"semantic_template":"qf-function-limit-affine-delta/{\"inputs\":[\"7\",\"1/2\",\"1/14\"],\"operation\":\"affine_delta_validity\"}"},"fields":{"title":"Affine δ Witness with Strict Punctured Interior","stem":"For arbitrary real A,L, define f(x)=L+(-7)(x−A) for x≠A on R∖{A}. With ε=2/4>0 and δ=2/28>0, does 0<|x−A|<δ guarantee |f(x)−L|<ε?","answer":{"kind":"short_text","value":"Yes","explanation":"Check |m|δ≤ε by exact cross products, giving Yes. Equality is permitted because the punctured input distance is strictly below δ. If m=0 every positive δ works. The target is L, not an undefined f(A)."},"skills":["Affine δ Witness with Strict Punctured Interior"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0059","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"14971c43faf44f1ca0186ecb26ee0d6484ee3b5f193fe9dec005047a8eee8a31","family_id":"qf-function-limit-affine-delta","seed":202610082059,"semantic_template":"qf-function-limit-affine-delta/{\"inputs\":[\"8\",\"3/1\",\"3/16\"],\"operation\":\"affine_delta_validity\"}"},"fields":{"title":"Affine δ Witness with Strict Punctured Interior","stem":"For arbitrary real A,L, define f(x)=L+(8)(x−A) for x≠A on R∖{A}. With ε=3/1>0 and δ=3/16>0, does 0<|x−A|<δ guarantee |f(x)−L|<ε?","answer":{"kind":"short_text","value":"Yes","explanation":"Check |m|δ≤ε by exact cross products, giving Yes. Equality is permitted because the punctured input distance is strictly below δ. If m=0 every positive δ works. The target is L, not an undefined f(A)."},"skills":["Affine δ Witness with Strict Punctured Interior"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0060","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2f2087492fc6095aa67be9665b62258752d91c2905e08a4a10cbf7a7de7fa296","family_id":"qf-function-limit-affine-delta","seed":202610082060,"semantic_template":"qf-function-limit-affine-delta/{\"inputs\":[\"9\",\"1/2\",\"1/9\"],\"operation\":\"affine_delta_validity\"}"},"fields":{"title":"Affine δ Witness with Strict Punctured Interior","stem":"For arbitrary real A,L, define f(x)=L+(-9)(x−A) for x≠A on R∖{A}. With ε=1/2>0 and δ=2/18>0, does 0<|x−A|<δ guarantee |f(x)−L|<ε?","answer":{"kind":"short_text","value":"No","explanation":"Check |m|δ≤ε by exact cross products, giving No. Equality is permitted because the punctured input distance is strictly below δ. If m=0 every positive δ works. The target is L, not an undefined f(A)."},"skills":["Affine δ Witness with Strict Punctured Interior"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0061","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"61473c9f270c35c84f56f5cc576312b0a3335705c0b15fce350a1da1579d0a9a","family_id":"qf-function-limit-affine-delta","seed":202610082061,"semantic_template":"qf-function-limit-affine-delta/{\"inputs\":[\"10\",\"2/3\",\"1/15\"],\"operation\":\"affine_delta_validity\"}"},"fields":{"title":"Affine δ Witness with Strict Punctured Interior","stem":"For arbitrary real A,L, define f(x)=L+(10)(x−A) for x≠A on R∖{A}. With ε=2/3>0 and δ=2/30>0, does 0<|x−A|<δ guarantee |f(x)−L|<ε?","answer":{"kind":"short_text","value":"Yes","explanation":"Check |m|δ≤ε by exact cross products, giving Yes. Equality is permitted because the punctured input distance is strictly below δ. If m=0 every positive δ works. The target is L, not an undefined f(A)."},"skills":["Affine δ Witness with Strict Punctured Interior"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0062","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"361c62e34900da09b344b7f7998b15774a54ca6ebfbd27108f1911fb0ff1668f","family_id":"qf-function-limit-affine-delta","seed":202610082062,"semantic_template":"qf-function-limit-affine-delta/{\"inputs\":[\"11\",\"3/4\",\"3/88\"],\"operation\":\"affine_delta_validity\"}"},"fields":{"title":"Affine δ Witness with Strict Punctured Interior","stem":"For arbitrary real A,L, define f(x)=L+(-11)(x−A) for x≠A on R∖{A}. With ε=3/4>0 and δ=3/88>0, does 0<|x−A|<δ guarantee |f(x)−L|<ε?","answer":{"kind":"short_text","value":"Yes","explanation":"Check |m|δ≤ε by exact cross products, giving Yes. Equality is permitted because the punctured input distance is strictly below δ. If m=0 every positive δ works. The target is L, not an undefined f(A)."},"skills":["Affine δ Witness with Strict Punctured Interior"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0063","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0c4ae5271e869b3475382b91924b8282fc7d43463bba200fef66836aff942c6c","family_id":"qf-function-limit-affine-delta","seed":202610082063,"semantic_template":"qf-function-limit-affine-delta/{\"inputs\":[\"12\",\"1/1\",\"1/6\"],\"operation\":\"affine_delta_validity\"}"},"fields":{"title":"Affine δ Witness with Strict Punctured Interior","stem":"For arbitrary real A,L, define f(x)=L+(12)(x−A) for x≠A on R∖{A}. With ε=1/1>0 and δ=2/12>0, does 0<|x−A|<δ guarantee |f(x)−L|<ε?","answer":{"kind":"short_text","value":"No","explanation":"Check |m|δ≤ε by exact cross products, giving No. Equality is permitted because the punctured input distance is strictly below δ. If m=0 every positive δ works. The target is L, not an undefined f(A)."},"skills":["Affine δ Witness with Strict Punctured Interior"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0064","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"05f3ce3fd8aaf6f7b85834db13510687b51227da3b0215b53720c4236d5937ec","family_id":"qf-function-limit-affine-delta","seed":202610082064,"semantic_template":"qf-function-limit-affine-delta/{\"inputs\":[\"13\",\"1/1\",\"1/13\"],\"operation\":\"affine_delta_validity\"}"},"fields":{"title":"Affine δ Witness with Strict Punctured Interior","stem":"For arbitrary real A,L, define f(x)=L+(-13)(x−A) for x≠A on R∖{A}. With ε=2/2>0 and δ=2/26>0, does 0<|x−A|<δ guarantee |f(x)−L|<ε?","answer":{"kind":"short_text","value":"Yes","explanation":"Check |m|δ≤ε by exact cross products, giving Yes. Equality is permitted because the punctured input distance is strictly below δ. If m=0 every positive δ works. The target is L, not an undefined f(A)."},"skills":["Affine δ Witness with Strict Punctured Interior"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0065","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"26d91d32fd7f48dd10a80ec2018020ed0b912c57be5bc72993f5e64541bf59b0","family_id":"qf-function-limit-affine-delta","seed":202610082065,"semantic_template":"qf-function-limit-affine-delta/{\"inputs\":[\"14\",\"1/1\",\"1/28\"],\"operation\":\"affine_delta_validity\"}"},"fields":{"title":"Affine δ Witness with Strict Punctured Interior","stem":"For arbitrary real A,L, define f(x)=L+(14)(x−A) for x≠A on R∖{A}. With ε=3/3>0 and δ=3/84>0, does 0<|x−A|<δ guarantee |f(x)−L|<ε?","answer":{"kind":"short_text","value":"Yes","explanation":"Check |m|δ≤ε by exact cross products, giving Yes. Equality is permitted because the punctured input distance is strictly below δ. If m=0 every positive δ works. The target is L, not an undefined f(A)."},"skills":["Affine δ Witness with Strict Punctured Interior"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0066","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"53afc3549b4ad359fe7ffb377783d11ad1f042f3dc052762113192f2861fd2c0","family_id":"qf-function-limit-affine-delta","seed":202610082066,"semantic_template":"qf-function-limit-affine-delta/{\"inputs\":[\"15\",\"1/4\",\"1/30\"],\"operation\":\"affine_delta_validity\"}"},"fields":{"title":"Affine δ Witness with Strict Punctured Interior","stem":"For arbitrary real A,L, define f(x)=L+(-15)(x−A) for x≠A on R∖{A}. With ε=1/4>0 and δ=2/60>0, does 0<|x−A|<δ guarantee |f(x)−L|<ε?","answer":{"kind":"short_text","value":"No","explanation":"Check |m|δ≤ε by exact cross products, giving No. Equality is permitted because the punctured input distance is strictly below δ. If m=0 every positive δ works. The target is L, not an undefined f(A)."},"skills":["Affine δ Witness with Strict Punctured Interior"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0067","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bf7051e7ba3904d30ba63f051958a1dd0a8920748c63bb932145a5900eaec618","family_id":"qf-function-limit-affine-delta","seed":202610082067,"semantic_template":"qf-function-limit-affine-delta/{\"inputs\":[\"16\",\"2/1\",\"1/8\"],\"operation\":\"affine_delta_validity\"}"},"fields":{"title":"Affine δ Witness with Strict Punctured Interior","stem":"For arbitrary real A,L, define f(x)=L+(16)(x−A) for x≠A on R∖{A}. With ε=2/1>0 and δ=2/16>0, does 0<|x−A|<δ guarantee |f(x)−L|<ε?","answer":{"kind":"short_text","value":"Yes","explanation":"Check |m|δ≤ε by exact cross products, giving Yes. Equality is permitted because the punctured input distance is strictly below δ. If m=0 every positive δ works. The target is L, not an undefined f(A)."},"skills":["Affine δ Witness with Strict Punctured Interior"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0068","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ea5d1977067a0277021f53baad10e7a10821b186f5a63f5fe6d9d03fd59bed2b","family_id":"qf-function-limit-affine-delta","seed":202610082068,"semantic_template":"qf-function-limit-affine-delta/{\"inputs\":[\"17\",\"3/2\",\"3/68\"],\"operation\":\"affine_delta_validity\"}"},"fields":{"title":"Affine δ Witness with Strict Punctured Interior","stem":"For arbitrary real A,L, define f(x)=L+(-17)(x−A) for x≠A on R∖{A}. With ε=3/2>0 and δ=3/68>0, does 0<|x−A|<δ guarantee |f(x)−L|<ε?","answer":{"kind":"short_text","value":"Yes","explanation":"Check |m|δ≤ε by exact cross products, giving Yes. Equality is permitted because the punctured input distance is strictly below δ. If m=0 every positive δ works. The target is L, not an undefined f(A)."},"skills":["Affine δ Witness with Strict Punctured Interior"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0069","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"764659d90502415a68acda84d43620b5c97f3b31b2b8fa5b49cd50918bbef960","family_id":"qf-function-limit-affine-delta","seed":202610082069,"semantic_template":"qf-function-limit-affine-delta/{\"inputs\":[\"18\",\"1/3\",\"1/27\"],\"operation\":\"affine_delta_validity\"}"},"fields":{"title":"Affine δ Witness with Strict Punctured Interior","stem":"For arbitrary real A,L, define f(x)=L+(18)(x−A) for x≠A on R∖{A}. With ε=1/3>0 and δ=2/54>0, does 0<|x−A|<δ guarantee |f(x)−L|<ε?","answer":{"kind":"short_text","value":"No","explanation":"Check |m|δ≤ε by exact cross products, giving No. Equality is permitted because the punctured input distance is strictly below δ. If m=0 every positive δ works. The target is L, not an undefined f(A)."},"skills":["Affine δ Witness with Strict Punctured Interior"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0070","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0acd2987419c69c2d1f8610c55e2c9382faa064673767d70a921d092e26011d1","family_id":"qf-function-limit-affine-delta","seed":202610082070,"semantic_template":"qf-function-limit-affine-delta/{\"inputs\":[\"19\",\"1/2\",\"1/38\"],\"operation\":\"affine_delta_validity\"}"},"fields":{"title":"Affine δ Witness with Strict Punctured Interior","stem":"For arbitrary real A,L, define f(x)=L+(-19)(x−A) for x≠A on R∖{A}. With ε=2/4>0 and δ=2/76>0, does 0<|x−A|<δ guarantee |f(x)−L|<ε?","answer":{"kind":"short_text","value":"Yes","explanation":"Check |m|δ≤ε by exact cross products, giving Yes. Equality is permitted because the punctured input distance is strictly below δ. If m=0 every positive δ works. The target is L, not an undefined f(A)."},"skills":["Affine δ Witness with Strict Punctured Interior"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0071","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f489189a0aee1882cd794e3f58620b535a994d2239bb0fe8e9c37e3c14bc1fbb","family_id":"qf-function-limit-affine-delta","seed":202610082071,"semantic_template":"qf-function-limit-affine-delta/{\"inputs\":[\"20\",\"3/1\",\"3/40\"],\"operation\":\"affine_delta_validity\"}"},"fields":{"title":"Affine δ Witness with Strict Punctured Interior","stem":"For arbitrary real A,L, define f(x)=L+(20)(x−A) for x≠A on R∖{A}. With ε=3/1>0 and δ=3/40>0, does 0<|x−A|<δ guarantee |f(x)−L|<ε?","answer":{"kind":"short_text","value":"Yes","explanation":"Check |m|δ≤ε by exact cross products, giving Yes. Equality is permitted because the punctured input distance is strictly below δ. If m=0 every positive δ works. The target is L, not an undefined f(A)."},"skills":["Affine δ Witness with Strict Punctured Interior"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0072","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"898b11142d126112a88fd7aaa0d4186315d297539c7d2f6d71d10eaec925a658","family_id":"qf-function-limit-affine-delta","seed":202610082072,"semantic_template":"qf-function-limit-affine-delta/{\"inputs\":[\"21\",\"1/2\",\"1/21\"],\"operation\":\"affine_delta_validity\"}"},"fields":{"title":"Affine δ Witness with Strict Punctured Interior","stem":"For arbitrary real A,L, define f(x)=L+(-21)(x−A) for x≠A on R∖{A}. With ε=1/2>0 and δ=2/42>0, does 0<|x−A|<δ guarantee |f(x)−L|<ε?","answer":{"kind":"short_text","value":"No","explanation":"Check |m|δ≤ε by exact cross products, giving No. Equality is permitted because the punctured input distance is strictly below δ. If m=0 every positive δ works. The target is L, not an undefined f(A)."},"skills":["Affine δ Witness with Strict Punctured Interior"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0073","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bcd1801f7c4c4f9d786bd60fbdd24690b69b3778443d2c5903d741c4ab1e1b03","family_id":"qf-function-limit-affine-delta","seed":202610082073,"semantic_template":"qf-function-limit-affine-delta/{\"inputs\":[\"22\",\"2/3\",\"1/33\"],\"operation\":\"affine_delta_validity\"}"},"fields":{"title":"Affine δ Witness with Strict Punctured Interior","stem":"For arbitrary real A,L, define f(x)=L+(22)(x−A) for x≠A on R∖{A}. With ε=2/3>0 and δ=2/66>0, does 0<|x−A|<δ guarantee |f(x)−L|<ε?","answer":{"kind":"short_text","value":"Yes","explanation":"Check |m|δ≤ε by exact cross products, giving Yes. Equality is permitted because the punctured input distance is strictly below δ. If m=0 every positive δ works. The target is L, not an undefined f(A)."},"skills":["Affine δ Witness with Strict Punctured Interior"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0074","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"28200e695c3ff18694ff087b9d0916b99af8147782cc0c5bcd7c47bc7b5de09c","family_id":"qf-function-limit-affine-delta","seed":202610082074,"semantic_template":"qf-function-limit-affine-delta/{\"inputs\":[\"23\",\"3/4\",\"3/184\"],\"operation\":\"affine_delta_validity\"}"},"fields":{"title":"Affine δ Witness with Strict Punctured Interior","stem":"For arbitrary real A,L, define f(x)=L+(-23)(x−A) for x≠A on R∖{A}. With ε=3/4>0 and δ=3/184>0, does 0<|x−A|<δ guarantee |f(x)−L|<ε?","answer":{"kind":"short_text","value":"Yes","explanation":"Check |m|δ≤ε by exact cross products, giving Yes. Equality is permitted because the punctured input distance is strictly below δ. If m=0 every positive δ works. The target is L, not an undefined f(A)."},"skills":["Affine δ Witness with Strict Punctured Interior"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0075","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"49ad1565e0990d42fb3f69f88add2f45103fa5bdff83410930295dee3a624764","family_id":"qf-function-limit-affine-delta","seed":202610082075,"semantic_template":"qf-function-limit-affine-delta/{\"inputs\":[\"24\",\"1/1\",\"1/12\"],\"operation\":\"affine_delta_validity\"}"},"fields":{"title":"Affine δ Witness with Strict Punctured Interior","stem":"For arbitrary real A,L, define f(x)=L+(24)(x−A) for x≠A on R∖{A}. With ε=1/1>0 and δ=2/24>0, does 0<|x−A|<δ guarantee |f(x)−L|<ε?","answer":{"kind":"short_text","value":"No","explanation":"Check |m|δ≤ε by exact cross products, giving No. Equality is permitted because the punctured input distance is strictly below δ. If m=0 every positive δ works. The target is L, not an undefined f(A)."},"skills":["Affine δ Witness with Strict Punctured Interior"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0076","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5ab3924139bbbe15e47a33808cd5c8bbdd93df9e617ab2e693547433fb2f8af9","family_id":"qf-function-limit-two-sided-affine","seed":202610082076,"semantic_template":"qf-function-limit-two-sided-affine/{\"inputs\":[\"-12\",\"-12\"],\"operation\":\"two_sided_affine_limit\"}"},"fields":{"title":"Two-Sided Limit of Affine Branches","stem":"On the punctured domain R∖{-12}, The left branch at x<-12 is (1)x+(0); the right branch at x>-12 is (-1)x+(-24). Find the finite two-sided limit, or answer does not exist when the two side limits differ.","answer":{"kind":"short_text","value":"-12","explanation":"The left and right limits are [-12, -12]. They must agree, giving -12; nonexistence of a two-sided limit does not erase either one-sided limit."},"skills":["Two-Sided Limit of Affine Branches"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0077","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"907197a3baaa4d49161ff678bdc91f9ab3585f05e888b11ffe5bf0d4e25db7b0","family_id":"qf-function-limit-two-sided-affine","seed":202610082077,"semantic_template":"qf-function-limit-two-sided-affine/{\"inputs\":[\"-11\",\"2\"],\"operation\":\"two_sided_affine_limit\"}"},"fields":{"title":"Two-Sided Limit of Affine Branches","stem":"On the punctured domain R∖{-11}, The left branch at x<-11 is (2)x+(11); the right branch at x>-11 is (-2)x+(-20). Find the finite two-sided limit, or answer does not exist when the two side limits differ.","answer":{"kind":"short_text","value":"does not exist","explanation":"The left and right limits are [-11, 2]. They must agree, giving does not exist; nonexistence of a two-sided limit does not erase either one-sided limit."},"skills":["Two-Sided Limit of Affine Branches"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0078","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b0f83ec5149c052afbd52fa229f3de1101108447682424439488e8eb6e2b8d61","family_id":"qf-function-limit-two-sided-affine","seed":202610082078,"semantic_template":"qf-function-limit-two-sided-affine/{\"inputs\":[\"-10\",\"3\"],\"operation\":\"two_sided_affine_limit\"}"},"fields":{"title":"Two-Sided Limit of Affine Branches","stem":"On the punctured domain R∖{-10}, The left branch at x<-10 is (3)x+(20); the right branch at x>-10 is (-3)x+(-27). Find the finite two-sided limit, or answer does not exist when the two side limits differ.","answer":{"kind":"short_text","value":"does not exist","explanation":"The left and right limits are [-10, 3]. They must agree, giving does not exist; nonexistence of a two-sided limit does not erase either one-sided limit."},"skills":["Two-Sided Limit of Affine Branches"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0079","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a4ad28115f0afda863bb6d8ae36d6f33dffc6990ddd756f2509b978391a73b73","family_id":"qf-function-limit-two-sided-affine","seed":202610082079,"semantic_template":"qf-function-limit-two-sided-affine/{\"inputs\":[\"-9\",\"-9\"],\"operation\":\"two_sided_affine_limit\"}"},"fields":{"title":"Two-Sided Limit of Affine Branches","stem":"On the punctured domain R∖{-9}, The left branch at x<-9 is (4)x+(27); the right branch at x>-9 is (-1)x+(-18). Find the finite two-sided limit, or answer does not exist when the two side limits differ.","answer":{"kind":"short_text","value":"-9","explanation":"The left and right limits are [-9, -9]. They must agree, giving -9; nonexistence of a two-sided limit does not erase either one-sided limit."},"skills":["Two-Sided Limit of Affine Branches"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0080","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"44fc9094d098680defdaf00d48dcda28394f6f6f235bb4a3a03389eb2f46404c","family_id":"qf-function-limit-two-sided-affine","seed":202610082080,"semantic_template":"qf-function-limit-two-sided-affine/{\"inputs\":[\"-8\",\"5\"],\"operation\":\"two_sided_affine_limit\"}"},"fields":{"title":"Two-Sided Limit of Affine Branches","stem":"On the punctured domain R∖{-8}, The left branch at x<-8 is (1)x+(0); the right branch at x>-8 is (-2)x+(-11). Find the finite two-sided limit, or answer does not exist when the two side limits differ.","answer":{"kind":"short_text","value":"does not exist","explanation":"The left and right limits are [-8, 5]. They must agree, giving does not exist; nonexistence of a two-sided limit does not erase either one-sided limit."},"skills":["Two-Sided Limit of Affine Branches"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0081","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3964c5e978060736f575aa6f0b7e3e1ef3dc8c50ec14352847edfe0187be840a","family_id":"qf-function-limit-two-sided-affine","seed":202610082081,"semantic_template":"qf-function-limit-two-sided-affine/{\"inputs\":[\"-7\",\"6\"],\"operation\":\"two_sided_affine_limit\"}"},"fields":{"title":"Two-Sided Limit of Affine Branches","stem":"On the punctured domain R∖{-7}, The left branch at x<-7 is (2)x+(7); the right branch at x>-7 is (-3)x+(-15). Find the finite two-sided limit, or answer does not exist when the two side limits differ.","answer":{"kind":"short_text","value":"does not exist","explanation":"The left and right limits are [-7, 6]. They must agree, giving does not exist; nonexistence of a two-sided limit does not erase either one-sided limit."},"skills":["Two-Sided Limit of Affine Branches"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0082","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"34fb0b387ff63bd4fbca2f37f012d6daecd65f5a3873bdec0f0b1959357ab92e","family_id":"qf-function-limit-two-sided-affine","seed":202610082082,"semantic_template":"qf-function-limit-two-sided-affine/{\"inputs\":[\"-6\",\"-6\"],\"operation\":\"two_sided_affine_limit\"}"},"fields":{"title":"Two-Sided Limit of Affine Branches","stem":"On the punctured domain R∖{-6}, The left branch at x<-6 is (3)x+(12); the right branch at x>-6 is (-1)x+(-12). Find the finite two-sided limit, or answer does not exist when the two side limits differ.","answer":{"kind":"short_text","value":"-6","explanation":"The left and right limits are [-6, -6]. They must agree, giving -6; nonexistence of a two-sided limit does not erase either one-sided limit."},"skills":["Two-Sided Limit of Affine Branches"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0083","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a02fe95595616693777e42a396c3be0c51fbec429272ed2b09f3e1da77cf706d","family_id":"qf-function-limit-two-sided-affine","seed":202610082083,"semantic_template":"qf-function-limit-two-sided-affine/{\"inputs\":[\"-5\",\"8\"],\"operation\":\"two_sided_affine_limit\"}"},"fields":{"title":"Two-Sided Limit of Affine Branches","stem":"On the punctured domain R∖{-5}, The left branch at x<-5 is (4)x+(15); the right branch at x>-5 is (-2)x+(-2). Find the finite two-sided limit, or answer does not exist when the two side limits differ.","answer":{"kind":"short_text","value":"does not exist","explanation":"The left and right limits are [-5, 8]. They must agree, giving does not exist; nonexistence of a two-sided limit does not erase either one-sided limit."},"skills":["Two-Sided Limit of Affine Branches"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0084","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"62d071a2f90cbf707cbd15aeb425a27452e655125f60a27ef2184c75cfa9579f","family_id":"qf-function-limit-two-sided-affine","seed":202610082084,"semantic_template":"qf-function-limit-two-sided-affine/{\"inputs\":[\"-4\",\"9\"],\"operation\":\"two_sided_affine_limit\"}"},"fields":{"title":"Two-Sided Limit of Affine Branches","stem":"On the punctured domain R∖{-4}, The left branch at x<-4 is (1)x+(0); the right branch at x>-4 is (-3)x+(-3). Find the finite two-sided limit, or answer does not exist when the two side limits differ.","answer":{"kind":"short_text","value":"does not exist","explanation":"The left and right limits are [-4, 9]. They must agree, giving does not exist; nonexistence of a two-sided limit does not erase either one-sided limit."},"skills":["Two-Sided Limit of Affine Branches"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0085","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8a99daef99a43352093fbb397bf3b0009413eeb820f7196370bd1df3e5404a99","family_id":"qf-function-limit-two-sided-affine","seed":202610082085,"semantic_template":"qf-function-limit-two-sided-affine/{\"inputs\":[\"-3\",\"-3\"],\"operation\":\"two_sided_affine_limit\"}"},"fields":{"title":"Two-Sided Limit of Affine Branches","stem":"On the punctured domain R∖{-3}, The left branch at x<-3 is (2)x+(3); the right branch at x>-3 is (-1)x+(-6). Find the finite two-sided limit, or answer does not exist when the two side limits differ.","answer":{"kind":"short_text","value":"-3","explanation":"The left and right limits are [-3, -3]. They must agree, giving -3; nonexistence of a two-sided limit does not erase either one-sided limit."},"skills":["Two-Sided Limit of Affine Branches"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0086","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"db333381c8e7311a17c3c3755e1fbec4b9b6714ea59611104d38a33a6eb585fb","family_id":"qf-function-limit-two-sided-affine","seed":202610082086,"semantic_template":"qf-function-limit-two-sided-affine/{\"inputs\":[\"-2\",\"11\"],\"operation\":\"two_sided_affine_limit\"}"},"fields":{"title":"Two-Sided Limit of Affine Branches","stem":"On the punctured domain R∖{-2}, The left branch at x<-2 is (3)x+(4); the right branch at x>-2 is (-2)x+(7). Find the finite two-sided limit, or answer does not exist when the two side limits differ.","answer":{"kind":"short_text","value":"does not exist","explanation":"The left and right limits are [-2, 11]. They must agree, giving does not exist; nonexistence of a two-sided limit does not erase either one-sided limit."},"skills":["Two-Sided Limit of Affine Branches"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0087","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8c14416789cf16e97bb8b24d6bd8e4a67c5e84f955be1c78a2de0961d428daa4","family_id":"qf-function-limit-two-sided-affine","seed":202610082087,"semantic_template":"qf-function-limit-two-sided-affine/{\"inputs\":[\"-1\",\"12\"],\"operation\":\"two_sided_affine_limit\"}"},"fields":{"title":"Two-Sided Limit of Affine Branches","stem":"On the punctured domain R∖{-1}, The left branch at x<-1 is (4)x+(3); the right branch at x>-1 is (-3)x+(9). Find the finite two-sided limit, or answer does not exist when the two side limits differ.","answer":{"kind":"short_text","value":"does not exist","explanation":"The left and right limits are [-1, 12]. They must agree, giving does not exist; nonexistence of a two-sided limit does not erase either one-sided limit."},"skills":["Two-Sided Limit of Affine Branches"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0088","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4fbe494b1cf487727d5243dc311e40ee2050ff909e72d2713a575e1094cc1e64","family_id":"qf-function-limit-two-sided-affine","seed":202610082088,"semantic_template":"qf-function-limit-two-sided-affine/{\"inputs\":[\"0\",\"0\"],\"operation\":\"two_sided_affine_limit\"}"},"fields":{"title":"Two-Sided Limit of Affine Branches","stem":"On the punctured domain R∖{0}, The left branch at x<0 is (1)x+(0); the right branch at x>0 is (-1)x+(0). Find the finite two-sided limit, or answer does not exist when the two side limits differ.","answer":{"kind":"short_text","value":"0","explanation":"The left and right limits are [0, 0]. They must agree, giving 0; nonexistence of a two-sided limit does not erase either one-sided limit."},"skills":["Two-Sided Limit of Affine Branches"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0089","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b4c42da070a641d60674af95d2666bbfa11760ffcfb2f490640c9f2c0dc1e406","family_id":"qf-function-limit-two-sided-affine","seed":202610082089,"semantic_template":"qf-function-limit-two-sided-affine/{\"inputs\":[\"1\",\"14\"],\"operation\":\"two_sided_affine_limit\"}"},"fields":{"title":"Two-Sided Limit of Affine Branches","stem":"On the punctured domain R∖{1}, The left branch at x<1 is (2)x+(-1); the right branch at x>1 is (-2)x+(16). Find the finite two-sided limit, or answer does not exist when the two side limits differ.","answer":{"kind":"short_text","value":"does not exist","explanation":"The left and right limits are [1, 14]. They must agree, giving does not exist; nonexistence of a two-sided limit does not erase either one-sided limit."},"skills":["Two-Sided Limit of Affine Branches"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0090","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f5570a961140adb5475830bae92aabe0517714c0c483b55b9921b524dbc80b75","family_id":"qf-function-limit-two-sided-affine","seed":202610082090,"semantic_template":"qf-function-limit-two-sided-affine/{\"inputs\":[\"2\",\"15\"],\"operation\":\"two_sided_affine_limit\"}"},"fields":{"title":"Two-Sided Limit of Affine Branches","stem":"On the punctured domain R∖{2}, The left branch at x<2 is (3)x+(-4); the right branch at x>2 is (-3)x+(21). Find the finite two-sided limit, or answer does not exist when the two side limits differ.","answer":{"kind":"short_text","value":"does not exist","explanation":"The left and right limits are [2, 15]. They must agree, giving does not exist; nonexistence of a two-sided limit does not erase either one-sided limit."},"skills":["Two-Sided Limit of Affine Branches"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0091","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4ae915b21f2b5ae68687eb6cebe91ecad59b4ff7e30a548dddb03e4176fa0ebc","family_id":"qf-function-limit-two-sided-affine","seed":202610082091,"semantic_template":"qf-function-limit-two-sided-affine/{\"inputs\":[\"3\",\"3\"],\"operation\":\"two_sided_affine_limit\"}"},"fields":{"title":"Two-Sided Limit of Affine Branches","stem":"On the punctured domain R∖{3}, The left branch at x<3 is (4)x+(-9); the right branch at x>3 is (-1)x+(6). Find the finite two-sided limit, or answer does not exist when the two side limits differ.","answer":{"kind":"short_text","value":"3","explanation":"The left and right limits are [3, 3]. They must agree, giving 3; nonexistence of a two-sided limit does not erase either one-sided limit."},"skills":["Two-Sided Limit of Affine Branches"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0092","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f9b48f80299e03e3e22372d9ce766f94469516b30ffed064bdf94909f0806dad","family_id":"qf-function-limit-two-sided-affine","seed":202610082092,"semantic_template":"qf-function-limit-two-sided-affine/{\"inputs\":[\"4\",\"17\"],\"operation\":\"two_sided_affine_limit\"}"},"fields":{"title":"Two-Sided Limit of Affine Branches","stem":"On the punctured domain R∖{4}, The left branch at x<4 is (1)x+(0); the right branch at x>4 is (-2)x+(25). Find the finite two-sided limit, or answer does not exist when the two side limits differ.","answer":{"kind":"short_text","value":"does not exist","explanation":"The left and right limits are [4, 17]. They must agree, giving does not exist; nonexistence of a two-sided limit does not erase either one-sided limit."},"skills":["Two-Sided Limit of Affine Branches"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0093","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5140e04c0e4b68b5f7daacb3f320f50a2becd7f22f0a8d645324ad2885b8755d","family_id":"qf-function-limit-two-sided-affine","seed":202610082093,"semantic_template":"qf-function-limit-two-sided-affine/{\"inputs\":[\"5\",\"18\"],\"operation\":\"two_sided_affine_limit\"}"},"fields":{"title":"Two-Sided Limit of Affine Branches","stem":"On the punctured domain R∖{5}, The left branch at x<5 is (2)x+(-5); the right branch at x>5 is (-3)x+(33). Find the finite two-sided limit, or answer does not exist when the two side limits differ.","answer":{"kind":"short_text","value":"does not exist","explanation":"The left and right limits are [5, 18]. They must agree, giving does not exist; nonexistence of a two-sided limit does not erase either one-sided limit."},"skills":["Two-Sided Limit of Affine Branches"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0094","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f7639744d6a40593dd884c26e3c3e62e884d26ed01a3003ac8ed833d51173fa4","family_id":"qf-function-limit-two-sided-affine","seed":202610082094,"semantic_template":"qf-function-limit-two-sided-affine/{\"inputs\":[\"6\",\"6\"],\"operation\":\"two_sided_affine_limit\"}"},"fields":{"title":"Two-Sided Limit of Affine Branches","stem":"On the punctured domain R∖{6}, The left branch at x<6 is (3)x+(-12); the right branch at x>6 is (-1)x+(12). Find the finite two-sided limit, or answer does not exist when the two side limits differ.","answer":{"kind":"short_text","value":"6","explanation":"The left and right limits are [6, 6]. They must agree, giving 6; nonexistence of a two-sided limit does not erase either one-sided limit."},"skills":["Two-Sided Limit of Affine Branches"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0095","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a2844d3dcfc05c8b33cf0379f4ba91f49d51559005778936b184ed8367ad14d3","family_id":"qf-function-limit-two-sided-affine","seed":202610082095,"semantic_template":"qf-function-limit-two-sided-affine/{\"inputs\":[\"7\",\"20\"],\"operation\":\"two_sided_affine_limit\"}"},"fields":{"title":"Two-Sided Limit of Affine Branches","stem":"On the punctured domain R∖{7}, The left branch at x<7 is (4)x+(-21); the right branch at x>7 is (-2)x+(34). Find the finite two-sided limit, or answer does not exist when the two side limits differ.","answer":{"kind":"short_text","value":"does not exist","explanation":"The left and right limits are [7, 20]. They must agree, giving does not exist; nonexistence of a two-sided limit does not erase either one-sided limit."},"skills":["Two-Sided Limit of Affine Branches"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0096","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0b156c2e708c7c2e90d812f5f70e15ea06e3485f3d660a6ef3e4f5c805e74d7b","family_id":"qf-function-limit-two-sided-affine","seed":202610082096,"semantic_template":"qf-function-limit-two-sided-affine/{\"inputs\":[\"8\",\"21\"],\"operation\":\"two_sided_affine_limit\"}"},"fields":{"title":"Two-Sided Limit of Affine Branches","stem":"On the punctured domain R∖{8}, The left branch at x<8 is (1)x+(0); the right branch at x>8 is (-3)x+(45). Find the finite two-sided limit, or answer does not exist when the two side limits differ.","answer":{"kind":"short_text","value":"does not exist","explanation":"The left and right limits are [8, 21]. They must agree, giving does not exist; nonexistence of a two-sided limit does not erase either one-sided limit."},"skills":["Two-Sided Limit of Affine Branches"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0097","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"002de0582362f1fac6d757d2ff9bb645dd683b09a6f1f2dd2b875c2eac435b8d","family_id":"qf-function-limit-two-sided-affine","seed":202610082097,"semantic_template":"qf-function-limit-two-sided-affine/{\"inputs\":[\"9\",\"9\"],\"operation\":\"two_sided_affine_limit\"}"},"fields":{"title":"Two-Sided Limit of Affine Branches","stem":"On the punctured domain R∖{9}, The left branch at x<9 is (2)x+(-9); the right branch at x>9 is (-1)x+(18). Find the finite two-sided limit, or answer does not exist when the two side limits differ.","answer":{"kind":"short_text","value":"9","explanation":"The left and right limits are [9, 9]. They must agree, giving 9; nonexistence of a two-sided limit does not erase either one-sided limit."},"skills":["Two-Sided Limit of Affine Branches"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0098","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e3ede648ba5701aa82da091fad703a064ecd92817f8b7190a87e4b2dd6ffb99c","family_id":"qf-function-limit-two-sided-affine","seed":202610082098,"semantic_template":"qf-function-limit-two-sided-affine/{\"inputs\":[\"10\",\"23\"],\"operation\":\"two_sided_affine_limit\"}"},"fields":{"title":"Two-Sided Limit of Affine Branches","stem":"On the punctured domain R∖{10}, The left branch at x<10 is (3)x+(-20); the right branch at x>10 is (-2)x+(43). Find the finite two-sided limit, or answer does not exist when the two side limits differ.","answer":{"kind":"short_text","value":"does not exist","explanation":"The left and right limits are [10, 23]. They must agree, giving does not exist; nonexistence of a two-sided limit does not erase either one-sided limit."},"skills":["Two-Sided Limit of Affine Branches"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0099","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4cc4df1b93dc6d5dae51bb23316e87b399fce2097c4da8fedb43e83f02f70f1c","family_id":"qf-function-limit-two-sided-affine","seed":202610082099,"semantic_template":"qf-function-limit-two-sided-affine/{\"inputs\":[\"11\",\"24\"],\"operation\":\"two_sided_affine_limit\"}"},"fields":{"title":"Two-Sided Limit of Affine Branches","stem":"On the punctured domain R∖{11}, The left branch at x<11 is (4)x+(-33); the right branch at x>11 is (-3)x+(57). Find the finite two-sided limit, or answer does not exist when the two side limits differ.","answer":{"kind":"short_text","value":"does not exist","explanation":"The left and right limits are [11, 24]. They must agree, giving does not exist; nonexistence of a two-sided limit does not erase either one-sided limit."},"skills":["Two-Sided Limit of Affine Branches"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r02-function-limit-0100","batch_id":"batch-20261007-advance-20261008-r02-function-limit-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"23a6932e8b10518deb4b9dcaf3f1a2435253462a02d6471a92aad486fd524d9d","family_id":"qf-function-limit-two-sided-affine","seed":202610082100,"semantic_template":"qf-function-limit-two-sided-affine/{\"inputs\":[\"12\",\"12\"],\"operation\":\"two_sided_affine_limit\"}"},"fields":{"title":"Two-Sided Limit of Affine Branches","stem":"On the punctured domain R∖{12}, The left branch at x<12 is (1)x+(0); the right branch at x>12 is (-1)x+(24). Find the finite two-sided limit, or answer does not exist when the two side limits differ.","answer":{"kind":"short_text","value":"12","explanation":"The left and right limits are [12, 12]. They must agree, giving 12; nonexistence of a two-sided limit does not erase either one-sided limit."},"skills":["Two-Sided Limit of Affine Branches"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}