{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-function-limit-01",
  "mko_id": "mko-function-limit",
  "title": {
    "zh-Hant": "函數極限：可去因子商、仿射 δ 界與不等的單側極限",
    "en": "Function limits: a removable-factor quotient, affine δ bounds and unequal one-sided limits"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-function-limit-01",
    "source": {
      "mko_id": "mko-function-limit",
      "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
      "sha256": "2dce49b3fcec9f9cf28bc85107ed2884a57b36ccc297c60e4ef5836b4a3c5ca3",
      "english_sha256": "608f319ceb9afeb2eb284c43c4a1d6d8d19c4e40dce81f0ac4df60cbb55f4c44"
    },
    "mode": "theorem_proof",
    "audience_prerequisites": [
      "abs_rules",
      "order_basics",
      "frac_rules",
      "distributive",
      "func_limit_def",
      "numeric_evaluation"
    ],
    "goal": {
      "zh-Hant": "在去心實數論域上：(1) 以 ε-δ 證明 f(x)=(x²−9)/(x−3) 滿足 lim_(x→3) f(x)=6，而 f(3) 仍無定義；(2) 對仿射 g(x)=4x−1 在 a=2、ε=1/2，證明 δ=1/8 合法（等號 |m|δ=ε 允許），δ=1/7 不合法；(3) 對分支函數 h(x)=x+1（x<1）、h(x)=2x+3（x>1），求左右極限 2 與 5，並證明兩側極限不存在。",
      "en": "On punctured real domains: (1) prove by ε-δ that f(x)=(x²−9)/(x−3) has lim_(x→3) f(x)=6 while f(3) stays undefined; (2) for the affine g(x)=4x−1 at a=2 with ε=1/2, prove that δ=1/8 is valid (the equality |m|δ=ε is allowed) and δ=1/7 is not; (3) for the branch function h(x)=x+1 (x<1), h(x)=2x+3 (x>1), find the one-sided limits 2 and 5 and prove that the two-sided limit does not exist."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "函數定義在去心論域 ℝ∖{a} 上；不要求 f(a) 存在。",
          "en": "Functions are defined on the punctured domain ℝ∖{a}; f(a) need not exist."
        }
      },
      {
        "id": "G2",
        "statement": {
          "zh-Hant": "f(x)=(x²−9)/(x−3)，論域 ℝ∖{3}。",
          "en": "f(x)=(x²−9)/(x−3) on ℝ∖{3}."
        }
      },
      {
        "id": "G3",
        "statement": {
          "zh-Hant": "g(x)=4x−1（m=4、b=−1），a=2；ε=1/2（p=1、q=2）；候選 δ=1/8（u=1、v=8）與 δ=1/7。",
          "en": "g(x)=4x−1 (m=4, b=−1), a=2; ε=1/2 (p=1, q=2); candidate δ=1/8 (u=1, v=8) and δ=1/7."
        }
      },
      {
        "id": "G4",
        "statement": {
          "zh-Hant": "h(x)=x+1（x<1），h(x)=2x+3（x>1），論域 ℝ∖{1}。",
          "en": "h(x)=x+1 for x<1 and h(x)=2x+3 for x>1, on ℝ∖{1}."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "G1",
          "func_limit_def"
        ],
        "before_tex": "\\lim_{x\\to a}f(x)=L",
        "after_tex": "\\forall\\varepsilon>0\\ \\exists\\delta>0\\ \\forall x:\\ 0<|x-a|<\\delta\\Rightarrow|f(x)-L|<\\varepsilon",
        "relation": "definition",
        "rule": "mko_func_limit",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依 MKO 展開去心極限。條件 0<|x−a|<δ 排除 x=a，所以 f(a) 是否存在、取什麼值都不進入定義；輸入距離是嚴格小於 δ，輸出誤差也是嚴格小於 ε。",
          "en": "Unfold the punctured limit by the MKO. The condition 0<|x−a|<δ excludes x=a, so whether f(a) exists, and its value, never enters the definition; the input distance is strictly below δ and the output error strictly below ε."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "mko_func_limit",
            "func_limit_def"
          ],
          "explanation": {
            "zh-Hant": "此步只展開定義，不推出新事實。",
            "en": "This step only unfolds the definition and derives no new fact."
          }
        },
        "leaf_basis": "func_limit_def + def_unfold"
      },
      {
        "id": "S02",
        "uses": [
          "G2",
          "S01"
        ],
        "before_tex": "\\mathrm{dom}(f)=\\mathbb{R}\\setminus\\{3\\}",
        "after_tex": "\\forall\\delta>0:\\ \\{x:0<|x-3|<\\delta\\}\\subseteq\\mathrm{dom}(f),\\ 3+\\tfrac{\\delta}{2}\\ \\text{in it}",
        "relation": "implies",
        "rule": "mko_func_limit",
        "conditions": [],
        "explanation": {
          "zh-Hant": "去心鄰域中的 x 都滿足 x≠3，所以都在 f 的論域內；而且 3+δ/2 屬於它（0<δ/2<δ），鄰域不空。因此「x→3 時 f 的極限」是有意義的問題，雖然 f(3)=(9−9)/(3−3) 是 0/0，沒有定義。",
          "en": "Every x in a punctured neighbourhood has x≠3, so it lies in the domain of f; and 3+δ/2 belongs to it (0<δ/2<δ), so the neighbourhood is not empty. Hence 'the limit of f as x→3' is a meaningful question, even though f(3)=(9−9)/(3−3) is 0/0 and undefined."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G2",
            "S01",
            "order_basics"
          ],
          "explanation": {
            "zh-Hant": "只確認極限問題有意義（3 是論域的極限點）；不宣稱極限存在。",
            "en": "It only confirms that the limit question is meaningful (3 is a limit point of the domain); it does not claim the limit exists."
          }
        },
        "leaf_basis": "set_builder + order_basics"
      },
      {
        "id": "S03",
        "uses": [
          "G2",
          "distributive"
        ],
        "before_tex": "x^2-9",
        "after_tex": "(x-3)(x+3)",
        "relation": "equivalent",
        "rule": "distributive",
        "conditions": [],
        "explanation": {
          "zh-Hant": "以分配律展開 (x−3)(x+3)=x²+3x−3x−9=x²−9，這是對所有實數 x 成立的恆等式（N1）。",
          "en": "Distributivity expands (x−3)(x+3)=x²+3x−3x−9=x²−9, an identity for every real x (N1)."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "distributive"
          ],
          "explanation": {
            "zh-Hant": "恆等式，雙向成立。",
            "en": "An identity, valid both ways."
          }
        },
        "leaf_basis": "distributive + add_inverse + add_zero"
      },
      {
        "id": "S04",
        "uses": [
          "S03",
          "frac_rules"
        ],
        "before_tex": "f(x)=\\frac{(x-3)(x+3)}{x-3}\\ (x\\neq 3)",
        "after_tex": "f(x)=x+3\\ (x\\neq 3)",
        "relation": "equivalent",
        "rule": "frac_rules",
        "conditions": [
          "x-3\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "對 x≠3，x−3≠0，可以約去公因子：(x+3)(x−3)/(1·(x−3))=(x+3)/1（a/b=(ak)/(bk)，k=x−3≠0）。這只在論域 ℝ∖{3} 上成立；它不給 f 在 3 的值，因為在 3 約分要除以 0。",
          "en": "For x≠3, x−3≠0, so the common factor cancels: (x+3)(x−3)/(1·(x−3))=(x+3)/1 (a/b=(ak)/(bk) with k=x−3≠0). This holds only on the domain ℝ∖{3}; it gives f no value at 3, since cancelling there would divide by 0."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "frac_rules",
            "mul_inverse_nonzero"
          ],
          "explanation": {
            "zh-Hant": "反向是分子分母同乘 x−3，需要 x−3≠0；條件在論域內處處成立。",
            "en": "Backward, multiply numerator and denominator by x−3, which needs x−3≠0; the condition holds throughout the domain."
          }
        },
        "leaf_basis": "frac_rules + mul_inverse_nonzero"
      },
      {
        "id": "S05",
        "uses": [
          "S04"
        ],
        "before_tex": "|f(x)-6|\\ (x\\neq 3)",
        "after_tex": "|(x+3)-6|=|x-3|",
        "relation": "equivalent",
        "rule": "eq_replace",
        "conditions": [
          "x\\neq 3"
        ],
        "explanation": {
          "zh-Hant": "在 x≠3 處以 S04 代入 f(x)=x+3，(x+3)−6=x−3。輸出誤差恰等於輸入距離。",
          "en": "At x≠3 substitute f(x)=x+3 from S04; (x+3)−6=x−3. The output error equals the input distance exactly."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "S04",
            "eq_replace"
          ],
          "explanation": {
            "zh-Hant": "等值代換，雙向成立；只在 x≠3 使用。",
            "en": "Replacement of equals, valid both ways; used only at x≠3."
          }
        },
        "leaf_basis": "eq_replace + add_assoc + add_inverse"
      },
      {
        "id": "S06",
        "uses": [
          "S01",
          "S02",
          "S05"
        ],
        "before_tex": "\\varepsilon>0\\ \\text{arbitrary},\\ \\delta=\\varepsilon",
        "after_tex": "0<|x-3|<\\delta\\Rightarrow|f(x)-6|=|x-3|<\\varepsilon;\\ \\lim_{x\\to 3}f(x)=6",
        "relation": "implies",
        "rule": "forall_intro",
        "conditions": [
          "\\varepsilon>0"
        ],
        "explanation": {
          "zh-Hant": "取任意 ε>0，令 δ=ε>0。若 0<|x−3|<δ，則 x≠3，由 S05，|f(x)−6|=|x−3|<δ=ε。ε 任意，依全稱推廣及 S01，lim_(x→3) f(x)=6。極限 6 不是 f(3)：f(3) 依然無定義（S02）。",
          "en": "Take an arbitrary ε>0 and set δ=ε>0. If 0<|x−3|<δ then x≠3, and by S05 |f(x)−6|=|x−3|<δ=ε. As ε is arbitrary, generalization and S01 give lim_(x→3) f(x)=6. The limit 6 is not f(3): f(3) is still undefined (S02)."
        },
        "reverse_audit": {
          "kind": "sufficient_goal_reduction",
          "required_ids": [
            "S01",
            "S05",
            "forall_intro"
          ],
          "explanation": {
            "zh-Hant": "目標 G：極限為 6。足夠條件 H：對任意 ε，δ=ε 有效；H⇒G 由定義。H 由 S05 正向證成，未把 G 當作前提，也沒有用到 f(3)。",
            "en": "Goal G: the limit is 6. Sufficient condition H: for arbitrary ε, δ=ε works; H⇒G by the definition. H is proved forwards from S05, without assuming G and without using f(3)."
          }
        },
        "leaf_basis": "forall_intro + exists_intro + func_limit_def"
      },
      {
        "id": "S07",
        "uses": [
          "G3",
          "abs_rules"
        ],
        "before_tex": "|g(x)-7|,\\ g(x)=4x-1,\\ 7=4\\cdot 2-1",
        "after_tex": "|4x-8|=|4(x-2)|=4|x-2|",
        "relation": "equivalent",
        "rule": "abs_rules",
        "conditions": [],
        "explanation": {
          "zh-Hant": "仿射極限值 ma+b=4·2−1=7（MKO 假設 a2）。g(x)−7=4x−8=4(x−2)（N3），依 |uv|=|u||v|，|g(x)−7|=4|x−2|。",
          "en": "The affine limit value is ma+b=4·2−1=7 (MKO assumption a2). g(x)−7=4x−8=4(x−2) (N3), and by |uv|=|u||v|, |g(x)−7|=4|x−2|."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "abs_rules",
            "distributive"
          ],
          "explanation": {
            "zh-Hant": "恆等變形，雙向成立。",
            "en": "Identities, valid both ways."
          }
        },
        "leaf_basis": "abs_rules + distributive + numeric_evaluation"
      },
      {
        "id": "S08",
        "uses": [
          "G3",
          "frac_rules"
        ],
        "before_tex": "|m|\\cdot\\frac{u}{v}\\le\\frac{p}{q}",
        "after_tex": "|m|\\,u\\,q\\le p\\,v",
        "relation": "equivalent",
        "rule": "order_basics",
        "conditions": [
          "q>0",
          "v>0"
        ],
        "explanation": {
          "zh-Hant": "兩邊同乘正數 qv，不等號方向不變，得整數判準 |m|uq≤pv，即 MKO 假設 a3 的形式。這裡是「≤」，等號允許。",
          "en": "Multiply both sides by the positive qv; the direction is unchanged, giving the integer criterion |m|uq≤pv, the form of MKO assumption a3. Here it is '≤': equality is allowed."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "order_basics",
            "frac_rules"
          ],
          "explanation": {
            "zh-Hant": "反向是同除以正數 qv。",
            "en": "Backward, divide by the positive qv."
          }
        },
        "leaf_basis": "order_basics + frac_rules"
      },
      {
        "id": "S09",
        "uses": [
          "S07",
          "S08",
          "G3"
        ],
        "before_tex": "4\\cdot 1\\cdot 2=8\\le 8=1\\cdot 8",
        "after_tex": "0<|x-2|<\\tfrac18\\Rightarrow|g(x)-7|<\\tfrac12",
        "relation": "theorem_application",
        "rule": "L_affine_bound",
        "conditions": [
          "|m|\\delta\\le\\varepsilon"
        ],
        "explanation": {
          "zh-Hant": "代入 m=4、u=1、v=8、p=1、q=2：|m|uq=8≤8=pv，等號成立，即 |m|δ=4·(1/8)=1/2=ε。依引理 L_affine_bound，δ=1/8 合法。等號可以，因為輸入距離嚴格小於 δ：4|x−2|<4·(1/8)=1/2。引理在 S09.1 至 S09.3 證明。",
          "en": "Substitute m=4, u=1, v=8, p=1, q=2: |m|uq=8≤8=pv, with equality, that is |m|δ=4·(1/8)=1/2=ε. By Lemma L_affine_bound, δ=1/8 is valid. Equality is fine because the input distance is strictly below δ: 4|x−2|<4·(1/8)=1/2. The lemma is proved in S09.1 to S09.3."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S08",
            "L_affine_bound",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "引理的前提 |m|δ≤ε 以 N4、N5 核對；引理只給出充分條件。",
            "en": "The lemma's hypothesis |m|δ≤ε is checked by N4 and N5; the lemma gives only a sufficient condition."
          }
        },
        "leaf_basis": "abs_rules + order_basics + case_analysis"
      },
      {
        "id": "S09.1",
        "uses": [
          "distributive",
          "abs_rules"
        ],
        "before_tex": "|(mx+b)-(ma+b)|",
        "after_tex": "|m(x-a)|=|m|\\,|x-a|",
        "relation": "equivalent",
        "rule": "abs_rules",
        "conditions": [],
        "explanation": {
          "zh-Hant": "(mx+b)−(ma+b)=mx−ma=m(x−a)（N13），再用 |uv|=|u||v|。對任何 m、a、b、x 成立。",
          "en": "(mx+b)−(ma+b)=mx−ma=m(x−a) (N13), then |uv|=|u||v|. This holds for every m, a, b, x."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "distributive",
            "abs_rules"
          ],
          "explanation": {
            "zh-Hant": "恆等式，雙向成立。",
            "en": "An identity, valid both ways."
          }
        },
        "leaf_basis": "distributive + abs_rules + add_inverse"
      },
      {
        "id": "S09.2",
        "uses": [
          "S09.1",
          "mul_zero"
        ],
        "before_tex": "m=0",
        "after_tex": "|g(x)-(ma+b)|=0<\\varepsilon\\ \\text{for every}\\ x",
        "relation": "implies",
        "rule": "mul_zero",
        "conditions": [
          "m=0"
        ],
        "explanation": {
          "zh-Hant": "情況 m=0：由 S09.1，誤差為 0·|x−a|=0<ε，對任何 x 成立，所以任何正 δ 都合法（MKO 假設 a3）。",
          "en": "Case m=0: by S09.1 the error is 0·|x−a|=0<ε for every x, so every positive δ works (MKO assumption a3)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S09.1",
            "mul_zero"
          ],
          "explanation": {
            "zh-Hant": "只需 ε>0；這是引理的第一個情況。",
            "en": "Only ε>0 is needed; this is the first case of the lemma."
          }
        },
        "leaf_basis": "mul_zero + order_basics"
      },
      {
        "id": "S09.3",
        "uses": [
          "S09.1",
          "order_basics"
        ],
        "before_tex": "m\\neq 0,\\ 0<|x-a|<\\delta,\\ |m|\\delta\\le\\varepsilon",
        "after_tex": "|m||x-a|<|m|\\delta\\le\\varepsilon",
        "relation": "implies",
        "rule": "order_basics",
        "conditions": [
          "|m|>0"
        ],
        "explanation": {
          "zh-Hant": "情況 m≠0：|m|>0，把嚴格不等式 |x−a|<δ 兩邊乘正數 |m|，得 |m||x−a|<|m|δ；再接上 |m|δ≤ε，依序的傳遞得嚴格的 |g(x)−(ma+b)|<ε。嚴格性來自輸入的 <δ，所以 |m|δ=ε 的等號情形也合法。兩個情況涵蓋所有 m，S09.1 至 S09.3 證得引理 L_affine_bound。",
          "en": "Case m≠0: |m|>0, so multiplying the strict |x−a|<δ by the positive |m| gives |m||x−a|<|m|δ; with |m|δ≤ε, transitivity of order gives the strict |g(x)−(ma+b)|<ε. The strictness comes from the input's <δ, so the equality case |m|δ=ε is also valid. The two cases cover every m, and S09.1 to S09.3 prove Lemma L_affine_bound."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S09.1",
            "order_basics"
          ],
          "explanation": {
            "zh-Hant": "需要 |m|>0 才能保持嚴格不等號；m=0 已在 S09.2 另行處理。",
            "en": "|m|>0 is needed to keep the strict inequality; m=0 is handled separately in S09.2."
          }
        },
        "leaf_basis": "order_basics + case_analysis"
      },
      {
        "id": "S10",
        "uses": [
          "S07",
          "S08",
          "G3"
        ],
        "before_tex": "\\delta=\\tfrac17:\\ 4\\cdot 1\\cdot 2=8>7=1\\cdot 7",
        "after_tex": "x=2+\\tfrac18:\\ 0<|x-2|<\\tfrac17,\\ |g(x)-7|=\\tfrac12\\not<\\tfrac12",
        "relation": "implies",
        "rule": "quantifier_negation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "δ=1/7 時判準 8≤7 不成立。這時引理不適用，但還要證明 δ=1/7 真的不合法：取 x=2+1/8，|x−2|=1/8<1/7 且 x≠2，而 |g(x)−7|=4·(1/8)=1/2，不小於 ε=1/2。一個反例就否定了「所有 0<|x−2|<1/7 的 x 都合格」。",
          "en": "For δ=1/7 the criterion 8≤7 fails. The lemma then does not apply, and it must still be shown that δ=1/7 really fails: take x=2+1/8; then |x−2|=1/8<1/7 and x≠2, while |g(x)−7|=4·(1/8)=1/2, not below ε=1/2. One counterexample refutes 'every x with 0<|x−2|<1/7 qualifies'."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S07",
            "quantifier_negation",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "判準失敗本身只說明引理不適用；不合法的結論來自具體反例（N6、N7、N8）。",
            "en": "The failed criterion only means the lemma does not apply; invalidity comes from the explicit counterexample (N6, N7, N8)."
          }
        },
        "leaf_basis": "quantifier_negation + numeric_evaluation + order_basics"
      },
      {
        "id": "S11",
        "uses": [
          "G4",
          "one_sided_def"
        ],
        "before_tex": "x<1:\\ |h(x)-2|",
        "after_tex": "|(x+1)-2|=|x-1|;\\ \\delta=\\varepsilon\\ \\text{gives}\\ L^-=2",
        "relation": "implies",
        "rule": "one_sided_def",
        "conditions": [
          "x<1"
        ],
        "explanation": {
          "zh-Hant": "左側 x<1 時 h(x)=x+1，(x+1)−2=x−1（N9）。對任意 ε>0 取 δ=ε：若 1−δ<x<1，則 |h(x)−2|=|x−1|<δ=ε。所以左極限是 2，即左分支 1·1+1 的值。",
          "en": "On the left, x<1, h(x)=x+1 and (x+1)−2=x−1 (N9). For arbitrary ε>0 take δ=ε: if 1−δ<x<1 then |h(x)−2|=|x−1|<δ=ε. So the left limit is 2, the value 1·1+1 of the left branch."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G4",
            "one_sided_def",
            "forall_intro"
          ],
          "explanation": {
            "zh-Hant": "只用左分支在 x<1 的公式；ε 任意。",
            "en": "Only the left branch formula on x<1 is used; ε is arbitrary."
          }
        },
        "leaf_basis": "one_sided_def + abs_rules + forall_intro"
      },
      {
        "id": "S12",
        "uses": [
          "G4",
          "one_sided_def"
        ],
        "before_tex": "x>1:\\ |h(x)-5|",
        "after_tex": "|(2x+3)-5|=2|x-1|;\\ \\delta=\\tfrac{\\varepsilon}{2}\\ \\text{gives}\\ L^+=5",
        "relation": "implies",
        "rule": "one_sided_def",
        "conditions": [
          "x>1"
        ],
        "explanation": {
          "zh-Hant": "右側 x>1 時 h(x)=2x+3，(2x+3)−5=2(x−1)（N10）。取 δ=ε/2：若 1<x<1+δ，則 |h(x)−5|=2|x−1|<2δ=ε。所以右極限是 5，即右分支 2·1+3 的值。",
          "en": "On the right, x>1, h(x)=2x+3 and (2x+3)−5=2(x−1) (N10). Take δ=ε/2: if 1<x<1+δ then |h(x)−5|=2|x−1|<2δ=ε. So the right limit is 5, the value 2·1+3 of the right branch."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G4",
            "one_sided_def",
            "forall_intro"
          ],
          "explanation": {
            "zh-Hant": "只用右分支在 x>1 的公式；兩個單側極限各自存在。",
            "en": "Only the right branch formula on x>1 is used; each one-sided limit exists on its own."
          }
        },
        "leaf_basis": "one_sided_def + abs_rules + forall_intro"
      },
      {
        "id": "S13",
        "uses": [
          "S01",
          "G4"
        ],
        "before_tex": "\\lim_{x\\to 1}h(x)=L\\ \\text{(assumed)},\\ \\varepsilon=\\tfrac32",
        "after_tex": "\\delta>0,\\ t=\\tfrac{\\delta}{2}:\\ |h(1+t)-L|<\\tfrac32,\\ |h(1-t)-L|<\\tfrac32",
        "relation": "implies",
        "rule": "exists_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "反證：假設兩側極限 L 存在。對 ε=3/2，定義給出某個 δ>0。令 t=δ/2，則 0<t<δ，點 1+t 與 1−t 都滿足 0<|x−1|<δ，所以 |h(1+t)−L|<3/2 且 |h(1−t)−L|<3/2。選 ε=3/2 是因為左右極限相差 3，3/2 是其一半。",
          "en": "By contradiction: suppose a two-sided limit L exists. For ε=3/2 the definition gives some δ>0. Let t=δ/2, so 0<t<δ and both points 1+t and 1−t satisfy 0<|x−1|<δ; hence |h(1+t)−L|<3/2 and |h(1−t)−L|<3/2. ε=3/2 is chosen because the one-sided limits differ by 3, and 3/2 is half of that."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S01",
            "exists_elim",
            "forall_elim"
          ],
          "explanation": {
            "zh-Hant": "δ 是由假設取出的新名字；之後的結論不含 δ。",
            "en": "δ is a fresh name taken from the assumption; the final conclusion does not mention δ."
          }
        },
        "leaf_basis": "exists_elim + forall_elim + order_basics"
      },
      {
        "id": "S14",
        "uses": [
          "S13",
          "abs_rules"
        ],
        "before_tex": "|h(1+t)-h(1-t)|\\le|h(1+t)-L|+|L-h(1-t)|<3",
        "after_tex": "h(1+t)-h(1-t)=3+3t>3:\\ \\text{contradiction};\\ \\lim_{x\\to 1}h(x)\\ \\text{does not exist}",
        "relation": "implies",
        "rule": "contradiction",
        "conditions": [
          "t>0"
        ],
        "explanation": {
          "zh-Hant": "三角不等式與 |L−h(1−t)|=|h(1−t)−L| 給出 |h(1+t)−h(1−t)|<3/2+3/2=3。但直接計算 h(1+t)−h(1−t)=(2(1+t)+3)−((1−t)+1)=3+3t（N11），因 t>0，它大於 3，絕對值也大於 3。矛盾，所以兩側極限不存在。",
          "en": "The triangle inequality and |L−h(1−t)|=|h(1−t)−L| give |h(1+t)−h(1−t)|<3/2+3/2=3. But directly h(1+t)−h(1−t)=(2(1+t)+3)−((1−t)+1)=3+3t (N11), which exceeds 3 since t>0, and so does its absolute value. Contradiction, so the two-sided limit does not exist."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S13",
            "abs_rules",
            "contradiction"
          ],
          "explanation": {
            "zh-Hant": "矛盾只否定了「兩側極限存在」這個假設；S11、S12 的單側極限不受影響。",
            "en": "The contradiction refutes only the assumption that a two-sided limit exists; the one-sided limits of S11 and S12 are unaffected."
          }
        },
        "leaf_basis": "contradiction + abs_rules + order_basics + numeric_evaluation"
      },
      {
        "id": "S15",
        "uses": [
          "S06",
          "S09",
          "S10",
          "S11",
          "S12",
          "S14"
        ],
        "before_tex": "\\text{S06, S09, S10, S11, S12, S14}",
        "after_tex": "\\lim_{x\\to3}f=6;\\ \\delta=\\tfrac18\\ \\text{valid},\\ \\tfrac17\\ \\text{not};\\ L^-=2\\neq 5=L^+,\\ \\text{no two-sided limit}",
        "relation": "implies",
        "rule": "and_intro_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "匯總三個結論。對 h，兩個單側極限 2 與 5 都合法存在，只是不相等，所以只否定兩側極限；不能把其中一側抹掉，也不能以 h(1)（無定義）來補。",
          "en": "Collect the three conclusions. For h, the one-sided limits 2 and 5 both exist legitimately but differ, so only the two-sided limit is denied; neither side may be erased, and h(1), which is undefined, cannot repair anything."
        },
        "reverse_audit": {
          "kind": "sufficient_goal_reduction",
          "required_ids": [
            "S06",
            "S09",
            "S10",
            "S11",
            "S12",
            "S14"
          ],
          "explanation": {
            "zh-Hant": "目標 G：本稿的三個結論。每一部分已由所列步驟正向證成；本步只合取，未把 G 當作前提。",
            "en": "Goal G: the three conclusions of this guide. Each part is proved forwards by the listed steps; this step only conjoins them, without assuming G."
          }
        },
        "leaf_basis": "and_intro_elim"
      }
    ],
    "result": {
      "zh-Hant": "f(x)=(x²−9)/(x−3) 在 ℝ∖{3} 上滿足 lim_(x→3) f(x)=6（δ=ε），而 f(3) 仍無定義。g(x)=4x−1 在 a=2、ε=1/2：δ=1/8 合法（4·1·2=8≤8，等號允許），δ=1/7 不合法（x=2+1/8 的誤差恰為 1/2）。h 的左極限為 2、右極限為 5，兩側極限不存在。",
      "en": "f(x)=(x²−9)/(x−3) on ℝ∖{3} has lim_(x→3) f(x)=6 (with δ=ε), while f(3) stays undefined. For g(x)=4x−1 at a=2 with ε=1/2: δ=1/8 is valid (4·1·2=8≤8, equality allowed) and δ=1/7 is not (at x=2+1/8 the error is exactly 1/2). h has left limit 2 and right limit 5, and no two-sided limit."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "f(3)=0/0 沒有定義，所以 lim_(x→3) f(x) 不存在",
          "en": "f(3)=0/0 is undefined, so lim_(x→3) f(x) does not exist"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "去心極限不看 f(3)；S06 證得極限為 6。",
          "en": "The punctured limit ignores f(3); S06 proves the limit is 6."
        }
      },
      {
        "claim": {
          "zh-Hant": "f(x)=x+3，所以 f(3)=6",
          "en": "f(x)=x+3, so f(3)=6"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "化簡只在 x≠3 成立；f 的論域是 ℝ∖{3}，在 3 約分要除以 0。",
          "en": "The simplification holds only at x≠3; the domain of f is ℝ∖{3}, and cancelling at 3 would divide by 0."
        }
      },
      {
        "claim": {
          "zh-Hant": "ε=1/2 時 δ=1/8 不合法，因為 |m|δ 恰等於 ε",
          "en": "For ε=1/2, δ=1/8 is invalid because |m|δ equals ε exactly"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "0<|x−2|<1/8 給出 4|x−2|<1/2，嚴格不等成立。",
          "en": "0<|x−2|<1/8 gives 4|x−2|<1/2, a strict inequality."
        }
      },
      {
        "claim": {
          "zh-Hant": "h 在 1 的兩側極限不存在，所以右極限也不存在",
          "en": "h has no two-sided limit at 1, so its right limit does not exist either"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "右極限為 5（S12）；只是不等於左極限 2。",
          "en": "The right limit is 5 (S12); it merely differs from the left limit 2."
        }
      },
      {
        "claim": {
          "zh-Hant": "f(3.1)=6.1、f(2.9)=5.9，所以極限為 6 已得證",
          "en": "f(3.1)=6.1 and f(2.9)=5.9, so the limit 6 is proved"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "有限測點只與結論相容，不是證明；證明是 S06 對任意 ε 的論證。",
          "en": "Finitely many test points are only consistent with the conclusion, not a proof; the proof is the argument of S06 for arbitrary ε."
        }
      }
    ],
    "status": "candidate",
    "gaps": [],
    "verification": {
      "structure": "checked_local_script",
      "mathematics": "candidate_argument",
      "formalization": "not_formalized"
    },
    "rule_glossary": [
      {
        "id": "abs_rules",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "絕對值（實數）：|u|=u（u≥0）、|u|=−u（u<0）；|u|≥0；|uv|=|u||v|；三角不等式 |u+v|≤|u|+|v|；|u|<r ⇔ −r<u<r。",
          "en": "Absolute value (reals): |u|=u if u≥0 and |u|=−u if u<0; |u|≥0; |uv|=|u||v|; triangle inequality |u+v|≤|u|+|v|; |u|<r ⇔ −r<u<r."
        }
      },
      {
        "id": "order_basics",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "實數的序：任兩數恰有 u<v、u=v、u>v 之一；u<v 且 v<w 推出 u<w；u<v 推出 u+w<v+w；u<v 且 w>0 推出 uw<vw；u<v 且 w<0 推出 uw>vw。",
          "en": "Order of the reals: exactly one of u<v, u=v, u>v holds; u<v and v<w give u<w; u<v gives u+w<v+w; u<v and w>0 give uw<vw; u<v and w<0 give uw>vw."
        }
      },
      {
        "id": "frac_rules",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "分數運算（分母皆非零）：a/c+b/c=(a+b)/c；(a/b)(c/d)=(ac)/(bd)；(a/b)/(c/d)=(ad)/(bc)（另需 c≠0）；a/b=(ak)/(bk)（k≠0）。",
          "en": "Fraction rules (all denominators nonzero): a/c+b/c=(a+b)/c; (a/b)(c/d)=(ac)/(bd); (a/b)/(c/d)=(ad)/(bc) (also needing c≠0); a/b=(ak)/(bk) for k≠0."
        }
      },
      {
        "id": "distributive",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "分配律：u(v+w)=uv+uw；配合乘法交換律也有 (v+w)u=vu+wu。",
          "en": "Distributivity: u(v+w)=uv+uw; with commutativity also (v+w)u=vu+wu."
        }
      },
      {
        "id": "func_limit_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "函數極限：lim_{x→a} f(x)=L 表示對每個 ε>0，存在 δ>0，使所有滿足 0<|x−a|<δ 的定義域點 x 都有 |f(x)−L|<ε（a 須為定義域的極限點）。",
          "en": "Limit of a function: lim_{x→a} f(x)=L means that for every ε>0 there is δ>0 such that |f(x)−L|<ε for every domain point x with 0<|x−a|<δ (a must be a limit point of the domain)."
        }
      },
      {
        "id": "numeric_evaluation",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "明確整數或有理數算式的精確求值；可以獨立重算。",
          "en": "Exact evaluation of an explicit integer or rational expression; independently recomputable."
        }
      },
      {
        "id": "def_unfold",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "展開定義：把一個已定義的詞換成它的定義內容（或反向收回），命題的真假不變。",
          "en": "Unfolding a definition: replacing a defined term by its definition (or folding it back) does not change whether a statement is true."
        }
      },
      {
        "id": "set_builder",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "集合建構式：x∈{y∈A ∣ P(y)} 當且僅當 x∈A 且 P(x)。",
          "en": "Set-builder notation: x∈{y∈A ∣ P(y)} if and only if x∈A and P(x)."
        }
      },
      {
        "id": "add_inverse",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "加法反元素：每個 u 都有 −u，使 u+(−u)=0；u−v 定義為 u+(−v)。",
          "en": "Additive inverse: every u has −u with u+(−u)=0; u−v means u+(−v)."
        }
      },
      {
        "id": "add_zero",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "加法單位元：u+0=u。",
          "en": "Additive identity: u+0=u."
        }
      },
      {
        "id": "mul_inverse_nonzero",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "非零數的倒數：在 ℚ、ℝ、ℂ 中，若 u≠0，存在 1/u 使 u·(1/u)=1；對 u≠0 定義 v/u 為 v·(1/u)。u=0 時沒有倒數，v/0 不定義。",
          "en": "Inverse of a nonzero number: in ℚ, ℝ or ℂ, if u≠0 there is 1/u with u·(1/u)=1, and v/u means v·(1/u). 0 has no inverse, so v/0 is undefined."
        }
      },
      {
        "id": "eq_replace",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "等值代換：命題中的一個式子可換成與它相等的式子，命題的真假不變。",
          "en": "Replacement of equals: an expression inside a statement may be replaced by an equal expression without changing whether the statement is true."
        }
      },
      {
        "id": "add_assoc",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "加法結合律：(u+v)+w=u+(v+w)。",
          "en": "Associativity of addition: (u+v)+w=u+(v+w)."
        }
      },
      {
        "id": "forall_intro",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "∀引入（全稱推廣）：若某命題對一個「任意的」元素 x 證成，而證明沒有對 x 加任何額外假設，則它對論域中所有 x 成立。",
          "en": "∀-introduction (generalization): if a statement is proved for an arbitrary element x, with no extra assumption on x, it holds for every x in the domain."
        }
      },
      {
        "id": "exists_intro",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "∃引入：若找到 a∈D 使 P(a) 成立，則 ∃x∈D, P(x)。",
          "en": "∃-introduction: if some a∈D satisfies P(a), then ∃x∈D, P(x)."
        }
      },
      {
        "id": "case_analysis",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "分情況論證：若 P∨Q 成立，且 P⇒R、Q⇒R 都已證，則 R 成立。所有情況必須涵蓋全部可能。",
          "en": "Proof by cases: if P∨Q holds and both P⇒R and Q⇒R are proved, then R holds. The cases must cover every possibility."
        }
      },
      {
        "id": "mul_zero",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "乘以零：u·0=0。理由：u·0=u·(0+0)=u·0+u·0，兩邊同減 u·0 得 0=u·0。",
          "en": "Multiplying by zero: u·0=0. Reason: u·0=u·(0+0)=u·0+u·0; subtracting u·0 from both sides gives 0=u·0."
        }
      },
      {
        "id": "quantifier_negation",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "量詞否定：¬(∀x, P(x)) 等價於 ∃x, ¬P(x)；¬(∃x, P(x)) 等價於 ∀x, ¬P(x)。",
          "en": "Negating quantifiers: ¬(∀x, P(x)) is equivalent to ∃x, ¬P(x), and ¬(∃x, P(x)) is equivalent to ∀x, ¬P(x)."
        }
      },
      {
        "id": "exists_elim",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "∃消去：由 ∃x∈D, P(x)，可取一個新名字 a 表示某個滿足 P(a) 的元素；只要最後結論不含 a，就可用 a 推理。",
          "en": "∃-elimination: from ∃x∈D, P(x) we may name some element a with P(a) by a fresh name; reasoning with a is allowed as long as the final conclusion does not mention a."
        }
      },
      {
        "id": "forall_elim",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "∀消去（代入特例）：若 ∀x∈D, P(x) 成立且 a∈D，則 P(a) 成立。",
          "en": "∀-elimination (instantiation): if ∀x∈D, P(x) holds and a∈D, then P(a) holds."
        }
      },
      {
        "id": "contradiction",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "反證法：若假設 ¬P 推出矛盾（某命題與其否定同時成立），則 P 成立。",
          "en": "Proof by contradiction: if assuming ¬P leads to a contradiction (a statement and its negation both hold), then P holds."
        }
      },
      {
        "id": "and_intro_elim",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "合取：由 P 與 Q 得 P∧Q；由 P∧Q 可分別得 P 與 Q。",
          "en": "Conjunction: from P and Q infer P∧Q; from P∧Q infer P and infer Q."
        }
      },
      {
        "id": "mko_func_limit",
        "kind": "cited",
        "source": {
          "mko_id": "mko-function-limit",
          "sha256": "2dce49b3fcec9f9cf28bc85107ed2884a57b36ccc297c60e4ef5836b4a3c5ca3",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "對在 a 附近所有 x≠a 都有定義的實函數 f，lim_(x→a) f(x)=L 指每個 ε>0 都有 δ>0，使 0<|x−a|<δ 時 |f(x)−L|<ε；左右有限極限必須相等才有兩側極限。仿射 mx+b 的極限為 ma+b；ε=p/q、δ=u/v 的仿射均勻界允許 |m|uq≤pv，等號合法。",
          "en": "For a real f defined at all x≠a near a, lim_(x→a) f(x)=L means each ε>0 admits δ>0 with |f(x)−L|<ε whenever 0<|x−a|<δ; a finite two-sided limit requires equal one-sided limits. The affine limit of mx+b is ma+b; for ε=p/q and δ=u/v the uniform affine bound permits |m|uq≤pv, equality included."
        }
      },
      {
        "id": "one_sided_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "左極限 L⁻：每個 ε>0 都有 δ>0，使 a−δ<x<a 時 |f(x)−L⁻|<ε；右極限 L⁺：a<x<a+δ 時 |f(x)−L⁺|<ε。",
          "en": "Left limit L⁻: each ε>0 admits δ>0 with |f(x)−L⁻|<ε whenever a−δ<x<a; right limit L⁺: the same with a<x<a+δ."
        }
      },
      {
        "id": "L_affine_bound",
        "kind": "lemma",
        "proved_in": [
          "S09.1",
          "S09.2",
          "S09.3"
        ],
        "statement": {
          "zh-Hant": "對 g(x)=mx+b、實數 a、ε>0、δ>0：若 |m|δ≤ε，則 0<|x−a|<δ 推出 |g(x)−(ma+b)|<ε。",
          "en": "For g(x)=mx+b, a real a, ε>0 and δ>0: if |m|δ≤ε, then 0<|x−a|<δ implies |g(x)−(ma+b)|<ε."
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "N1",
        "step": "S03",
        "vars": [
          "x"
        ],
        "degree": 2,
        "lhs": "(x-3)*(x+3)",
        "rhs": "x**2-9"
      },
      {
        "id": "N2",
        "step": "S04",
        "lhs": "(3.1**2-9)/(3.1-3)",
        "rhs": "3.1+3"
      },
      {
        "id": "N3",
        "step": "S07",
        "vars": [
          "x"
        ],
        "degree": 1,
        "lhs": "(4*x-1)-7",
        "rhs": "4*(x-2)"
      },
      {
        "id": "N4",
        "step": "S09",
        "lhs": "4*1*2",
        "relation": "<=",
        "rhs": "1*8"
      },
      {
        "id": "N5",
        "step": "S09",
        "lhs": "4*(1/8)",
        "rhs": "1/2"
      },
      {
        "id": "N6",
        "step": "S10",
        "lhs": "4*1*2",
        "relation": ">",
        "rhs": "1*7"
      },
      {
        "id": "N7",
        "step": "S10",
        "lhs": "1/8",
        "relation": "<",
        "rhs": "1/7"
      },
      {
        "id": "N8",
        "step": "S10",
        "lhs": "abs((4*(2+1/8)-1)-7)",
        "rhs": "1/2"
      },
      {
        "id": "N9",
        "step": "S11",
        "vars": [
          "x"
        ],
        "degree": 1,
        "lhs": "(x+1)-2",
        "rhs": "x-1"
      },
      {
        "id": "N10",
        "step": "S12",
        "vars": [
          "x"
        ],
        "degree": 1,
        "lhs": "(2*x+3)-5",
        "rhs": "2*(x-1)"
      },
      {
        "id": "N11",
        "step": "S14",
        "vars": [
          "t"
        ],
        "degree": 1,
        "lhs": "(2*(1+t)+3)-((1-t)+1)",
        "rhs": "3+3*t"
      },
      {
        "id": "N12",
        "step": "S14",
        "lhs": "3/2+3/2",
        "rhs": "3"
      },
      {
        "id": "N13",
        "step": "S09.1",
        "vars": [
          "m",
          "x",
          "a",
          "b"
        ],
        "degree": 2,
        "lhs": "(m*x+b)-(m*a+b)",
        "rhs": "m*(x-a)"
      },
      {
        "id": "N14",
        "lhs": "(2.9**2-9)/(2.9-3)",
        "rhs": "5.9"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "只處理 MKO 的三類題：仿射式、可去因子商、左右仿射分支，論域為 ℝ∖{a}。不證明一般極限定律（和、積、合成），不處理孤立點或其他接近論域。不改動 canonical MKO 或英文 edition。",
      "en": "Treats only the MKO's three kinds of exercise: affine expressions, removable-factor quotients and left/right affine branches, on the domain ℝ∖{a}. It does not prove general limit laws (sums, products, composition) and does not treat isolated points or other approach domains. It does not change the canonical MKO or its English edition."
    },
    "md_symbols": [
      [
        "\\lim_{x\\to a}f(x)=L",
        "去心兩側極限",
        "the punctured two-sided limit"
      ],
      [
        "\\varepsilon,\\ \\delta",
        "輸出容差與輸入距離界，皆為正",
        "the output tolerance and the input distance bound, both positive"
      ],
      [
        "0<|x-a|<\\delta",
        "去心鄰域：x≠a 且距離嚴格小於 δ",
        "the punctured neighbourhood: x≠a and distance strictly below δ"
      ],
      [
        "\\varepsilon=\\tfrac{p}{q},\\ \\delta=\\tfrac{u}{v}",
        "有理容差與有理距離界",
        "rational tolerance and rational distance bound"
      ],
      [
        "L^-,\\ L^+",
        "左極限與右極限",
        "the left and right limits"
      ]
    ],
    "md_error_controls": [
      {
        "title": "分清極限與 f(a)，不能除零後補值 / Distinguish the limit from f(a); do not patch a value in after division by zero",
        "zh": "MKO 的誤解之一是把極限當 f(a)。f(3) 是 0/0，沒有定義；極限 6 來自 x≠3 處的化簡 f(x)=x+3（S04），而 0<|x−3| 把中心排除在外（S01）。化簡不等於在 3 定義了 f；若另外定義 f(3)=100，極限仍是 6。",
        "en": "One misconception the MKO lists is taking the limit to be f(a). f(3) is 0/0 and undefined; the limit 6 comes from the simplification f(x)=x+3 at x≠3 (S04), and 0<|x−3| keeps the centre out (S01). The simplification does not define f at 3; if one separately set f(3)=100, the limit would still be 6."
      },
      {
        "title": "δ 邊界的等號合法 / Equality at the δ boundary is allowed",
        "zh": "另一個誤解是把 |m|δ=ε 當成不合法。因為輸入距離嚴格小於 δ，4|x−2|<4δ=ε 仍是嚴格的（S09.3），所以 δ=1/8 合法。這和數列尾界不同：那裡比較的是誤差本身，誤差等於 ε 不合格。δ=1/7 不合法則要用具體反例證明（S10）。",
        "en": "Another misconception is rejecting |m|δ=ε. Because the input distance is strictly below δ, 4|x−2|<4δ=ε stays strict (S09.3), so δ=1/8 is valid. This differs from sequence cutoffs, where the error itself is compared and an error equal to ε fails. That δ=1/7 fails must be shown by an explicit counterexample (S10)."
      },
      {
        "title": "單側極限不等只否定兩側極限 / Unequal one-sided limits deny only the two-sided limit",
        "zh": "h 的左右極限 2 與 5 都存在，只是不等，所以兩側極限不存在（S14）。這不表示右極限不存在，也不能用有限個測點（例如 x=0.9、1.1）代替 ε-δ 論證。",
        "en": "The one-sided limits 2 and 5 of h both exist but differ, so there is no two-sided limit (S14). This does not mean the right limit fails to exist, and finitely many test points (for example x=0.9 and 1.1) cannot replace the ε-δ argument."
      }
    ],
    "md_open_items": [
      {
        "zh-Hant": "本稿不證明一般極限定律，也不處理論域含孤立點的情形。",
        "en": "This guide does not prove general limit laws and does not treat domains with isolated points."
      },
      {
        "zh-Hant": "N2 是兩個測點的數值核對，只與 S04 的化簡相容，不參與證明。",
        "en": "N2 is a numerical check at two test points, consistent with the simplification of S04; it plays no part in the proof."
      }
    ]
  },
  "revision": {
    "id": "production-20261019-adopted-96",
    "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
    "adopted_source_sha256": "f75050de6a74ead986b8699e576b6dd6e5e0acae0be6646bdbb41b727c4ac99e",
    "original_source_sha256": "f75050de6a74ead986b8699e576b6dd6e5e0acae0be6646bdbb41b727c4ac99e",
    "owner_adopted": true,
    "bilingual_sync": "complete_source_bound",
    "formal_Lean": "NOT_RUN",
    "independent_AI_concurrence": "NOT_MEASURED",
    "issue_ids": [],
    "scope": "Owner-adopted natural-language teaching guide; not a formal proof, canonical MKO rewrite or evidence-status promotion."
  }
}