# Sequence Limits and Exact Tail Cutoff Foundations

This batch has passed Schema, deduplication, source-snapshot, and independent-answer recomputation checks; it is still not marked as human-reviewed.

This question bank has corresponding bilingual step-by-step derivations. Source candidate and English ai_candidate states are retained. AI review, finite computation and Lean formal proof are labelled separately.

batch-20261007-advance-20261008-r01-sequence-limit-foundations · 2026-10-07 · 100 questions

This English edition is an AI candidate and does not yet have an OCME Atlas (GPT-6) Twin closure receipt.



## 1. Limit of an Explicit Reciprocal-Tail Sequence

On natural indices starting at zero, u(n)=(-12)+(0)/(n+1), with d≥1. Find the declared reciprocal-tail limit.

View answer and explanation-12

The reciprocal error tends to zero in this model, leaving L=-12, including c=0. This is not an inference about an arbitrary sequence from finitely many terms.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0001/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-reciprocal-value · candidate





## 2. Limit of an Explicit Reciprocal-Tail Sequence

On natural indices starting at zero, u(n)=(-11)+(-2)/(n+2), with d≥1. Find the declared reciprocal-tail limit.

View answer and explanation-11

The reciprocal error tends to zero in this model, leaving L=-11, including c=0. This is not an inference about an arbitrary sequence from finitely many terms.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0002/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-reciprocal-value · candidate





## 3. Limit of an Explicit Reciprocal-Tail Sequence

On natural indices starting at zero, u(n)=(-10)+(3)/(n+3), with d≥1. Find the declared reciprocal-tail limit.

View answer and explanation-10

The reciprocal error tends to zero in this model, leaving L=-10, including c=0. This is not an inference about an arbitrary sequence from finitely many terms.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0003/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-reciprocal-value · candidate





## 4. Limit of an Explicit Reciprocal-Tail Sequence

On natural indices starting at zero, u(n)=(-9)+(-4)/(n+4), with d≥1. Find the declared reciprocal-tail limit.

View answer and explanation-9

The reciprocal error tends to zero in this model, leaving L=-9, including c=0. This is not an inference about an arbitrary sequence from finitely many terms.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0004/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-reciprocal-value · candidate





## 5. Limit of an Explicit Reciprocal-Tail Sequence

On natural indices starting at zero, u(n)=(-8)+(5)/(n+5), with d≥1. Find the declared reciprocal-tail limit.

View answer and explanation-8

The reciprocal error tends to zero in this model, leaving L=-8, including c=0. This is not an inference about an arbitrary sequence from finitely many terms.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0005/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-reciprocal-value · candidate





## 6. Limit of an Explicit Reciprocal-Tail Sequence

On natural indices starting at zero, u(n)=(-7)+(-6)/(n+1), with d≥1. Find the declared reciprocal-tail limit.

View answer and explanation-7

The reciprocal error tends to zero in this model, leaving L=-7, including c=0. This is not an inference about an arbitrary sequence from finitely many terms.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0006/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-reciprocal-value · candidate





## 7. Limit of an Explicit Reciprocal-Tail Sequence

On natural indices starting at zero, u(n)=(-6)+(7)/(n+2), with d≥1. Find the declared reciprocal-tail limit.

View answer and explanation-6

The reciprocal error tends to zero in this model, leaving L=-6, including c=0. This is not an inference about an arbitrary sequence from finitely many terms.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0007/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-reciprocal-value · candidate





## 8. Limit of an Explicit Reciprocal-Tail Sequence

On natural indices starting at zero, u(n)=(-5)+(-8)/(n+3), with d≥1. Find the declared reciprocal-tail limit.

View answer and explanation-5

The reciprocal error tends to zero in this model, leaving L=-5, including c=0. This is not an inference about an arbitrary sequence from finitely many terms.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0008/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-reciprocal-value · candidate





## 9. Limit of an Explicit Reciprocal-Tail Sequence

On natural indices starting at zero, u(n)=(-4)+(9)/(n+4), with d≥1. Find the declared reciprocal-tail limit.

View answer and explanation-4

The reciprocal error tends to zero in this model, leaving L=-4, including c=0. This is not an inference about an arbitrary sequence from finitely many terms.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0009/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-reciprocal-value · candidate





## 10. Limit of an Explicit Reciprocal-Tail Sequence

On natural indices starting at zero, u(n)=(-3)+(-10)/(n+5), with d≥1. Find the declared reciprocal-tail limit.

View answer and explanation-3

The reciprocal error tends to zero in this model, leaving L=-3, including c=0. This is not an inference about an arbitrary sequence from finitely many terms.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0010/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-reciprocal-value · candidate





## 11. Limit of an Explicit Reciprocal-Tail Sequence

On natural indices starting at zero, u(n)=(-2)+(11)/(n+1), with d≥1. Find the declared reciprocal-tail limit.

View answer and explanation-2

The reciprocal error tends to zero in this model, leaving L=-2, including c=0. This is not an inference about an arbitrary sequence from finitely many terms.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0011/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-reciprocal-value · candidate





## 12. Limit of an Explicit Reciprocal-Tail Sequence

On natural indices starting at zero, u(n)=(-1)+(-12)/(n+2), with d≥1. Find the declared reciprocal-tail limit.

View answer and explanation-1

The reciprocal error tends to zero in this model, leaving L=-1, including c=0. This is not an inference about an arbitrary sequence from finitely many terms.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0012/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-reciprocal-value · candidate





## 13. Limit of an Explicit Reciprocal-Tail Sequence

On natural indices starting at zero, u(n)=(0)+(13)/(n+3), with d≥1. Find the declared reciprocal-tail limit.

View answer and explanation0

The reciprocal error tends to zero in this model, leaving L=0, including c=0. This is not an inference about an arbitrary sequence from finitely many terms.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0013/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-reciprocal-value · candidate





## 14. Limit of an Explicit Reciprocal-Tail Sequence

On natural indices starting at zero, u(n)=(1)+(-14)/(n+4), with d≥1. Find the declared reciprocal-tail limit.

View answer and explanation1

The reciprocal error tends to zero in this model, leaving L=1, including c=0. This is not an inference about an arbitrary sequence from finitely many terms.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0014/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-reciprocal-value · candidate





## 15. Limit of an Explicit Reciprocal-Tail Sequence

On natural indices starting at zero, u(n)=(2)+(15)/(n+5), with d≥1. Find the declared reciprocal-tail limit.

View answer and explanation2

The reciprocal error tends to zero in this model, leaving L=2, including c=0. This is not an inference about an arbitrary sequence from finitely many terms.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0015/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-reciprocal-value · candidate





## 16. Limit of an Explicit Reciprocal-Tail Sequence

On natural indices starting at zero, u(n)=(3)+(-16)/(n+1), with d≥1. Find the declared reciprocal-tail limit.

View answer and explanation3

The reciprocal error tends to zero in this model, leaving L=3, including c=0. This is not an inference about an arbitrary sequence from finitely many terms.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0016/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-reciprocal-value · candidate





## 17. Limit of an Explicit Reciprocal-Tail Sequence

On natural indices starting at zero, u(n)=(4)+(17)/(n+2), with d≥1. Find the declared reciprocal-tail limit.

View answer and explanation4

The reciprocal error tends to zero in this model, leaving L=4, including c=0. This is not an inference about an arbitrary sequence from finitely many terms.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0017/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-reciprocal-value · candidate





## 18. Limit of an Explicit Reciprocal-Tail Sequence

On natural indices starting at zero, u(n)=(5)+(-18)/(n+3), with d≥1. Find the declared reciprocal-tail limit.

View answer and explanation5

The reciprocal error tends to zero in this model, leaving L=5, including c=0. This is not an inference about an arbitrary sequence from finitely many terms.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0018/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-reciprocal-value · candidate





## 19. Limit of an Explicit Reciprocal-Tail Sequence

On natural indices starting at zero, u(n)=(6)+(19)/(n+4), with d≥1. Find the declared reciprocal-tail limit.

View answer and explanation6

The reciprocal error tends to zero in this model, leaving L=6, including c=0. This is not an inference about an arbitrary sequence from finitely many terms.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0019/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-reciprocal-value · candidate





## 20. Limit of an Explicit Reciprocal-Tail Sequence

On natural indices starting at zero, u(n)=(7)+(-20)/(n+5), with d≥1. Find the declared reciprocal-tail limit.

View answer and explanation7

The reciprocal error tends to zero in this model, leaving L=7, including c=0. This is not an inference about an arbitrary sequence from finitely many terms.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0020/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-reciprocal-value · candidate





## 21. Limit of an Explicit Reciprocal-Tail Sequence

On natural indices starting at zero, u(n)=(8)+(21)/(n+1), with d≥1. Find the declared reciprocal-tail limit.

View answer and explanation8

The reciprocal error tends to zero in this model, leaving L=8, including c=0. This is not an inference about an arbitrary sequence from finitely many terms.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0021/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-reciprocal-value · candidate





## 22. Limit of an Explicit Reciprocal-Tail Sequence

On natural indices starting at zero, u(n)=(9)+(-22)/(n+2), with d≥1. Find the declared reciprocal-tail limit.

View answer and explanation9

The reciprocal error tends to zero in this model, leaving L=9, including c=0. This is not an inference about an arbitrary sequence from finitely many terms.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0022/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-reciprocal-value · candidate





## 23. Limit of an Explicit Reciprocal-Tail Sequence

On natural indices starting at zero, u(n)=(10)+(23)/(n+3), with d≥1. Find the declared reciprocal-tail limit.

View answer and explanation10

The reciprocal error tends to zero in this model, leaving L=10, including c=0. This is not an inference about an arbitrary sequence from finitely many terms.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0023/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-reciprocal-value · candidate





## 24. Limit of an Explicit Reciprocal-Tail Sequence

On natural indices starting at zero, u(n)=(11)+(-24)/(n+4), with d≥1. Find the declared reciprocal-tail limit.

View answer and explanation11

The reciprocal error tends to zero in this model, leaving L=11, including c=0. This is not an inference about an arbitrary sequence from finitely many terms.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0024/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-reciprocal-value · candidate





## 25. Limit of an Explicit Reciprocal-Tail Sequence

On natural indices starting at zero, u(n)=(12)+(25)/(n+5), with d≥1. Find the declared reciprocal-tail limit.

View answer and explanation12

The reciprocal error tends to zero in this model, leaving L=12, including c=0. This is not an inference about an arbitrary sequence from finitely many terms.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0025/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-reciprocal-value · candidate





## 26. Least Natural Cutoff Including the Cutoff Term

For an arbitrary fixed real L, u(n)=L+(0)/(n+1) on natural indices starting at zero, with d=1≥1. For ε=1/1>0, find the least natural N≥0 for which every n≥N satisfies |u(n)−L|<ε.

View answer and explanation0

The nonincreasing tail error reduces this model to |c|q<p(N+d). Its least cutoff is max(0,floor(|c|q/p)−d+1)=0. Equality fails the strict bound; a zero tail permits N=0.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0026/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-least-cutoff · candidate





## 27. Least Natural Cutoff Including the Cutoff Term

For an arbitrary fixed real L, u(n)=L+(-2)/(n+2) on natural indices starting at zero, with d=2≥1. For ε=2/2>0, find the least natural N≥0 for which every n≥N satisfies |u(n)−L|<ε.

View answer and explanation1

The nonincreasing tail error reduces this model to |c|q<p(N+d). Its least cutoff is max(0,floor(|c|q/p)−d+1)=1. Equality fails the strict bound; a zero tail permits N=0.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0027/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-least-cutoff · candidate





## 28. Least Natural Cutoff Including the Cutoff Term

For an arbitrary fixed real L, u(n)=L+(3)/(n+3) on natural indices starting at zero, with d=3≥1. For ε=3/3>0, find the least natural N≥0 for which every n≥N satisfies |u(n)−L|<ε.

View answer and explanation1

The nonincreasing tail error reduces this model to |c|q<p(N+d). Its least cutoff is max(0,floor(|c|q/p)−d+1)=1. Equality fails the strict bound; a zero tail permits N=0.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0028/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-least-cutoff · candidate





## 29. Least Natural Cutoff Including the Cutoff Term

For an arbitrary fixed real L, u(n)=L+(-4)/(n+4) on natural indices starting at zero, with d=4≥1. For ε=4/1>0, find the least natural N≥0 for which every n≥N satisfies |u(n)−L|<ε.

View answer and explanation0

The nonincreasing tail error reduces this model to |c|q<p(N+d). Its least cutoff is max(0,floor(|c|q/p)−d+1)=0. Equality fails the strict bound; a zero tail permits N=0.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0029/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-least-cutoff · candidate





## 30. Least Natural Cutoff Including the Cutoff Term

For an arbitrary fixed real L, u(n)=L+(5)/(n+1) on natural indices starting at zero, with d=1≥1. For ε=5/2>0, find the least natural N≥0 for which every n≥N satisfies |u(n)−L|<ε.

View answer and explanation2

The nonincreasing tail error reduces this model to |c|q<p(N+d). Its least cutoff is max(0,floor(|c|q/p)−d+1)=2. Equality fails the strict bound; a zero tail permits N=0.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0030/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-least-cutoff · candidate





## 31. Least Natural Cutoff Including the Cutoff Term

For an arbitrary fixed real L, u(n)=L+(-6)/(n+2) on natural indices starting at zero, with d=2≥1. For ε=1/3>0, find the least natural N≥0 for which every n≥N satisfies |u(n)−L|<ε.

View answer and explanation17

The nonincreasing tail error reduces this model to |c|q<p(N+d). Its least cutoff is max(0,floor(|c|q/p)−d+1)=17. Equality fails the strict bound; a zero tail permits N=0.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0031/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-least-cutoff · candidate





## 32. Least Natural Cutoff Including the Cutoff Term

For an arbitrary fixed real L, u(n)=L+(7)/(n+3) on natural indices starting at zero, with d=3≥1. For ε=2/1>0, find the least natural N≥0 for which every n≥N satisfies |u(n)−L|<ε.

View answer and explanation1

The nonincreasing tail error reduces this model to |c|q<p(N+d). Its least cutoff is max(0,floor(|c|q/p)−d+1)=1. Equality fails the strict bound; a zero tail permits N=0.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0032/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-least-cutoff · candidate





## 33. Least Natural Cutoff Including the Cutoff Term

For an arbitrary fixed real L, u(n)=L+(-8)/(n+4) on natural indices starting at zero, with d=4≥1. For ε=3/2>0, find the least natural N≥0 for which every n≥N satisfies |u(n)−L|<ε.

View answer and explanation2

The nonincreasing tail error reduces this model to |c|q<p(N+d). Its least cutoff is max(0,floor(|c|q/p)−d+1)=2. Equality fails the strict bound; a zero tail permits N=0.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0033/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-least-cutoff · candidate





## 34. Least Natural Cutoff Including the Cutoff Term

For an arbitrary fixed real L, u(n)=L+(9)/(n+1) on natural indices starting at zero, with d=1≥1. For ε=4/3>0, find the least natural N≥0 for which every n≥N satisfies |u(n)−L|<ε.

View answer and explanation6

The nonincreasing tail error reduces this model to |c|q<p(N+d). Its least cutoff is max(0,floor(|c|q/p)−d+1)=6. Equality fails the strict bound; a zero tail permits N=0.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0034/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-least-cutoff · candidate





## 35. Least Natural Cutoff Including the Cutoff Term

For an arbitrary fixed real L, u(n)=L+(-10)/(n+2) on natural indices starting at zero, with d=2≥1. For ε=5/1>0, find the least natural N≥0 for which every n≥N satisfies |u(n)−L|<ε.

View answer and explanation1

The nonincreasing tail error reduces this model to |c|q<p(N+d). Its least cutoff is max(0,floor(|c|q/p)−d+1)=1. Equality fails the strict bound; a zero tail permits N=0.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0035/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-least-cutoff · candidate





## 36. Least Natural Cutoff Including the Cutoff Term

For an arbitrary fixed real L, u(n)=L+(11)/(n+3) on natural indices starting at zero, with d=3≥1. For ε=1/2>0, find the least natural N≥0 for which every n≥N satisfies |u(n)−L|<ε.

View answer and explanation20

The nonincreasing tail error reduces this model to |c|q<p(N+d). Its least cutoff is max(0,floor(|c|q/p)−d+1)=20. Equality fails the strict bound; a zero tail permits N=0.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0036/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-least-cutoff · candidate





## 37. Least Natural Cutoff Including the Cutoff Term

For an arbitrary fixed real L, u(n)=L+(-12)/(n+4) on natural indices starting at zero, with d=4≥1. For ε=2/3>0, find the least natural N≥0 for which every n≥N satisfies |u(n)−L|<ε.

View answer and explanation15

The nonincreasing tail error reduces this model to |c|q<p(N+d). Its least cutoff is max(0,floor(|c|q/p)−d+1)=15. Equality fails the strict bound; a zero tail permits N=0.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0037/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-least-cutoff · candidate





## 38. Least Natural Cutoff Including the Cutoff Term

For an arbitrary fixed real L, u(n)=L+(13)/(n+1) on natural indices starting at zero, with d=1≥1. For ε=3/1>0, find the least natural N≥0 for which every n≥N satisfies |u(n)−L|<ε.

View answer and explanation4

The nonincreasing tail error reduces this model to |c|q<p(N+d). Its least cutoff is max(0,floor(|c|q/p)−d+1)=4. Equality fails the strict bound; a zero tail permits N=0.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0038/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-least-cutoff · candidate





## 39. Least Natural Cutoff Including the Cutoff Term

For an arbitrary fixed real L, u(n)=L+(-14)/(n+2) on natural indices starting at zero, with d=2≥1. For ε=4/2>0, find the least natural N≥0 for which every n≥N satisfies |u(n)−L|<ε.

View answer and explanation6

The nonincreasing tail error reduces this model to |c|q<p(N+d). Its least cutoff is max(0,floor(|c|q/p)−d+1)=6. Equality fails the strict bound; a zero tail permits N=0.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0039/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-least-cutoff · candidate





## 40. Least Natural Cutoff Including the Cutoff Term

For an arbitrary fixed real L, u(n)=L+(15)/(n+3) on natural indices starting at zero, with d=3≥1. For ε=5/3>0, find the least natural N≥0 for which every n≥N satisfies |u(n)−L|<ε.

View answer and explanation7

The nonincreasing tail error reduces this model to |c|q<p(N+d). Its least cutoff is max(0,floor(|c|q/p)−d+1)=7. Equality fails the strict bound; a zero tail permits N=0.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0040/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-least-cutoff · candidate





## 41. Least Natural Cutoff Including the Cutoff Term

For an arbitrary fixed real L, u(n)=L+(-16)/(n+4) on natural indices starting at zero, with d=4≥1. For ε=1/1>0, find the least natural N≥0 for which every n≥N satisfies |u(n)−L|<ε.

View answer and explanation13

The nonincreasing tail error reduces this model to |c|q<p(N+d). Its least cutoff is max(0,floor(|c|q/p)−d+1)=13. Equality fails the strict bound; a zero tail permits N=0.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0041/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-least-cutoff · candidate





## 42. Least Natural Cutoff Including the Cutoff Term

For an arbitrary fixed real L, u(n)=L+(17)/(n+1) on natural indices starting at zero, with d=1≥1. For ε=2/2>0, find the least natural N≥0 for which every n≥N satisfies |u(n)−L|<ε.

View answer and explanation17

The nonincreasing tail error reduces this model to |c|q<p(N+d). Its least cutoff is max(0,floor(|c|q/p)−d+1)=17. Equality fails the strict bound; a zero tail permits N=0.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0042/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-least-cutoff · candidate





## 43. Least Natural Cutoff Including the Cutoff Term

For an arbitrary fixed real L, u(n)=L+(-18)/(n+2) on natural indices starting at zero, with d=2≥1. For ε=3/3>0, find the least natural N≥0 for which every n≥N satisfies |u(n)−L|<ε.

View answer and explanation17

The nonincreasing tail error reduces this model to |c|q<p(N+d). Its least cutoff is max(0,floor(|c|q/p)−d+1)=17. Equality fails the strict bound; a zero tail permits N=0.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0043/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-least-cutoff · candidate





## 44. Least Natural Cutoff Including the Cutoff Term

For an arbitrary fixed real L, u(n)=L+(19)/(n+3) on natural indices starting at zero, with d=3≥1. For ε=4/1>0, find the least natural N≥0 for which every n≥N satisfies |u(n)−L|<ε.

View answer and explanation2

The nonincreasing tail error reduces this model to |c|q<p(N+d). Its least cutoff is max(0,floor(|c|q/p)−d+1)=2. Equality fails the strict bound; a zero tail permits N=0.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0044/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-least-cutoff · candidate





## 45. Least Natural Cutoff Including the Cutoff Term

For an arbitrary fixed real L, u(n)=L+(-20)/(n+4) on natural indices starting at zero, with d=4≥1. For ε=5/2>0, find the least natural N≥0 for which every n≥N satisfies |u(n)−L|<ε.

View answer and explanation5

The nonincreasing tail error reduces this model to |c|q<p(N+d). Its least cutoff is max(0,floor(|c|q/p)−d+1)=5. Equality fails the strict bound; a zero tail permits N=0.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0045/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-least-cutoff · candidate





## 46. Least Natural Cutoff Including the Cutoff Term

For an arbitrary fixed real L, u(n)=L+(21)/(n+1) on natural indices starting at zero, with d=1≥1. For ε=1/3>0, find the least natural N≥0 for which every n≥N satisfies |u(n)−L|<ε.

View answer and explanation63

The nonincreasing tail error reduces this model to |c|q<p(N+d). Its least cutoff is max(0,floor(|c|q/p)−d+1)=63. Equality fails the strict bound; a zero tail permits N=0.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0046/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-least-cutoff · candidate





## 47. Least Natural Cutoff Including the Cutoff Term

For an arbitrary fixed real L, u(n)=L+(-22)/(n+2) on natural indices starting at zero, with d=2≥1. For ε=2/1>0, find the least natural N≥0 for which every n≥N satisfies |u(n)−L|<ε.

View answer and explanation10

The nonincreasing tail error reduces this model to |c|q<p(N+d). Its least cutoff is max(0,floor(|c|q/p)−d+1)=10. Equality fails the strict bound; a zero tail permits N=0.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0047/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-least-cutoff · candidate





## 48. Least Natural Cutoff Including the Cutoff Term

For an arbitrary fixed real L, u(n)=L+(23)/(n+3) on natural indices starting at zero, with d=3≥1. For ε=3/2>0, find the least natural N≥0 for which every n≥N satisfies |u(n)−L|<ε.

View answer and explanation13

The nonincreasing tail error reduces this model to |c|q<p(N+d). Its least cutoff is max(0,floor(|c|q/p)−d+1)=13. Equality fails the strict bound; a zero tail permits N=0.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0048/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-least-cutoff · candidate





## 49. Least Natural Cutoff Including the Cutoff Term

For an arbitrary fixed real L, u(n)=L+(-24)/(n+4) on natural indices starting at zero, with d=4≥1. For ε=4/3>0, find the least natural N≥0 for which every n≥N satisfies |u(n)−L|<ε.

View answer and explanation15

The nonincreasing tail error reduces this model to |c|q<p(N+d). Its least cutoff is max(0,floor(|c|q/p)−d+1)=15. Equality fails the strict bound; a zero tail permits N=0.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0049/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-least-cutoff · candidate





## 50. Least Natural Cutoff Including the Cutoff Term

For an arbitrary fixed real L, u(n)=L+(25)/(n+1) on natural indices starting at zero, with d=1≥1. For ε=5/1>0, find the least natural N≥0 for which every n≥N satisfies |u(n)−L|<ε.

View answer and explanation5

The nonincreasing tail error reduces this model to |c|q<p(N+d). Its least cutoff is max(0,floor(|c|q/p)−d+1)=5. Equality fails the strict bound; a zero tail permits N=0.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0050/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-least-cutoff · candidate





## 51. Strict Eligibility of a Candidate Tail Cutoff

For an arbitrary fixed real L, u(n)=L+(0)/(n+1) on natural indices starting at zero, with d=1≥1. With ε=1/1>0 and candidate N=0, does every n≥N have |u(n)−L|<ε, including n=N?

View answer and explanationYes

Check |c|q<p(N+d) exactly, giving Yes. Equality is not strict enough. The all-tail reduction uses this model's nonincreasing error, not sampling as a proof for an arbitrary sequence.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0051/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-cutoff-validity · candidate





## 52. Strict Eligibility of a Candidate Tail Cutoff

For an arbitrary fixed real L, u(n)=L+(-2)/(n+2) on natural indices starting at zero, with d=2≥1. With ε=2/2>0 and candidate N=0, does every n≥N have |u(n)−L|<ε, including n=N?

View answer and explanationNo

Check |c|q<p(N+d) exactly, giving No. Equality is not strict enough. The all-tail reduction uses this model's nonincreasing error, not sampling as a proof for an arbitrary sequence.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0052/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-cutoff-validity · candidate





## 53. Strict Eligibility of a Candidate Tail Cutoff

For an arbitrary fixed real L, u(n)=L+(3)/(n+3) on natural indices starting at zero, with d=3≥1. With ε=3/3>0 and candidate N=3, does every n≥N have |u(n)−L|<ε, including n=N?

View answer and explanationYes

Check |c|q<p(N+d) exactly, giving Yes. Equality is not strict enough. The all-tail reduction uses this model's nonincreasing error, not sampling as a proof for an arbitrary sequence.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0053/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-cutoff-validity · candidate





## 54. Strict Eligibility of a Candidate Tail Cutoff

For an arbitrary fixed real L, u(n)=L+(-4)/(n+4) on natural indices starting at zero, with d=4≥1. With ε=4/1>0 and candidate N=0, does every n≥N have |u(n)−L|<ε, including n=N?

View answer and explanationYes

Check |c|q<p(N+d) exactly, giving Yes. Equality is not strict enough. The all-tail reduction uses this model's nonincreasing error, not sampling as a proof for an arbitrary sequence.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0054/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-cutoff-validity · candidate





## 55. Strict Eligibility of a Candidate Tail Cutoff

For an arbitrary fixed real L, u(n)=L+(5)/(n+1) on natural indices starting at zero, with d=1≥1. With ε=5/2>0 and candidate N=1, does every n≥N have |u(n)−L|<ε, including n=N?

View answer and explanationNo

Check |c|q<p(N+d) exactly, giving No. Equality is not strict enough. The all-tail reduction uses this model's nonincreasing error, not sampling as a proof for an arbitrary sequence.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0055/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-cutoff-validity · candidate





## 56. Strict Eligibility of a Candidate Tail Cutoff

For an arbitrary fixed real L, u(n)=L+(-6)/(n+2) on natural indices starting at zero, with d=2≥1. With ε=1/3>0 and candidate N=19, does every n≥N have |u(n)−L|<ε, including n=N?

View answer and explanationYes

Check |c|q<p(N+d) exactly, giving Yes. Equality is not strict enough. The all-tail reduction uses this model's nonincreasing error, not sampling as a proof for an arbitrary sequence.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0056/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-cutoff-validity · candidate





## 57. Strict Eligibility of a Candidate Tail Cutoff

For an arbitrary fixed real L, u(n)=L+(7)/(n+3) on natural indices starting at zero, with d=3≥1. With ε=2/1>0 and candidate N=1, does every n≥N have |u(n)−L|<ε, including n=N?

View answer and explanationYes

Check |c|q<p(N+d) exactly, giving Yes. Equality is not strict enough. The all-tail reduction uses this model's nonincreasing error, not sampling as a proof for an arbitrary sequence.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0057/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-cutoff-validity · candidate





## 58. Strict Eligibility of a Candidate Tail Cutoff

For an arbitrary fixed real L, u(n)=L+(-8)/(n+4) on natural indices starting at zero, with d=4≥1. With ε=3/2>0 and candidate N=1, does every n≥N have |u(n)−L|<ε, including n=N?

View answer and explanationNo

Check |c|q<p(N+d) exactly, giving No. Equality is not strict enough. The all-tail reduction uses this model's nonincreasing error, not sampling as a proof for an arbitrary sequence.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0058/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-cutoff-validity · candidate





## 59. Strict Eligibility of a Candidate Tail Cutoff

For an arbitrary fixed real L, u(n)=L+(9)/(n+1) on natural indices starting at zero, with d=1≥1. With ε=4/3>0 and candidate N=8, does every n≥N have |u(n)−L|<ε, including n=N?

View answer and explanationYes

Check |c|q<p(N+d) exactly, giving Yes. Equality is not strict enough. The all-tail reduction uses this model's nonincreasing error, not sampling as a proof for an arbitrary sequence.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0059/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-cutoff-validity · candidate





## 60. Strict Eligibility of a Candidate Tail Cutoff

For an arbitrary fixed real L, u(n)=L+(-10)/(n+2) on natural indices starting at zero, with d=2≥1. With ε=5/1>0 and candidate N=1, does every n≥N have |u(n)−L|<ε, including n=N?

View answer and explanationYes

Check |c|q<p(N+d) exactly, giving Yes. Equality is not strict enough. The all-tail reduction uses this model's nonincreasing error, not sampling as a proof for an arbitrary sequence.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0060/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-cutoff-validity · candidate





## 61. Strict Eligibility of a Candidate Tail Cutoff

For an arbitrary fixed real L, u(n)=L+(11)/(n+3) on natural indices starting at zero, with d=3≥1. With ε=1/2>0 and candidate N=19, does every n≥N have |u(n)−L|<ε, including n=N?

View answer and explanationNo

Check |c|q<p(N+d) exactly, giving No. Equality is not strict enough. The all-tail reduction uses this model's nonincreasing error, not sampling as a proof for an arbitrary sequence.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0061/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-cutoff-validity · candidate





## 62. Strict Eligibility of a Candidate Tail Cutoff

For an arbitrary fixed real L, u(n)=L+(-12)/(n+4) on natural indices starting at zero, with d=4≥1. With ε=2/3>0 and candidate N=17, does every n≥N have |u(n)−L|<ε, including n=N?

View answer and explanationYes

Check |c|q<p(N+d) exactly, giving Yes. Equality is not strict enough. The all-tail reduction uses this model's nonincreasing error, not sampling as a proof for an arbitrary sequence.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0062/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-cutoff-validity · candidate





## 63. Strict Eligibility of a Candidate Tail Cutoff

For an arbitrary fixed real L, u(n)=L+(13)/(n+1) on natural indices starting at zero, with d=1≥1. With ε=3/1>0 and candidate N=4, does every n≥N have |u(n)−L|<ε, including n=N?

View answer and explanationYes

Check |c|q<p(N+d) exactly, giving Yes. Equality is not strict enough. The all-tail reduction uses this model's nonincreasing error, not sampling as a proof for an arbitrary sequence.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0063/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-cutoff-validity · candidate





## 64. Strict Eligibility of a Candidate Tail Cutoff

For an arbitrary fixed real L, u(n)=L+(-14)/(n+2) on natural indices starting at zero, with d=2≥1. With ε=4/2>0 and candidate N=5, does every n≥N have |u(n)−L|<ε, including n=N?

View answer and explanationNo

Check |c|q<p(N+d) exactly, giving No. Equality is not strict enough. The all-tail reduction uses this model's nonincreasing error, not sampling as a proof for an arbitrary sequence.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0064/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-cutoff-validity · candidate





## 65. Strict Eligibility of a Candidate Tail Cutoff

For an arbitrary fixed real L, u(n)=L+(15)/(n+3) on natural indices starting at zero, with d=3≥1. With ε=5/3>0 and candidate N=9, does every n≥N have |u(n)−L|<ε, including n=N?

View answer and explanationYes

Check |c|q<p(N+d) exactly, giving Yes. Equality is not strict enough. The all-tail reduction uses this model's nonincreasing error, not sampling as a proof for an arbitrary sequence.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0065/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-cutoff-validity · candidate





## 66. Strict Eligibility of a Candidate Tail Cutoff

For an arbitrary fixed real L, u(n)=L+(-16)/(n+4) on natural indices starting at zero, with d=4≥1. With ε=1/1>0 and candidate N=13, does every n≥N have |u(n)−L|<ε, including n=N?

View answer and explanationYes

Check |c|q<p(N+d) exactly, giving Yes. Equality is not strict enough. The all-tail reduction uses this model's nonincreasing error, not sampling as a proof for an arbitrary sequence.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0066/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-cutoff-validity · candidate





## 67. Strict Eligibility of a Candidate Tail Cutoff

For an arbitrary fixed real L, u(n)=L+(17)/(n+1) on natural indices starting at zero, with d=1≥1. With ε=2/2>0 and candidate N=16, does every n≥N have |u(n)−L|<ε, including n=N?

View answer and explanationNo

Check |c|q<p(N+d) exactly, giving No. Equality is not strict enough. The all-tail reduction uses this model's nonincreasing error, not sampling as a proof for an arbitrary sequence.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0067/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-cutoff-validity · candidate





## 68. Strict Eligibility of a Candidate Tail Cutoff

For an arbitrary fixed real L, u(n)=L+(-18)/(n+2) on natural indices starting at zero, with d=2≥1. With ε=3/3>0 and candidate N=19, does every n≥N have |u(n)−L|<ε, including n=N?

View answer and explanationYes

Check |c|q<p(N+d) exactly, giving Yes. Equality is not strict enough. The all-tail reduction uses this model's nonincreasing error, not sampling as a proof for an arbitrary sequence.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0068/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-cutoff-validity · candidate





## 69. Strict Eligibility of a Candidate Tail Cutoff

For an arbitrary fixed real L, u(n)=L+(19)/(n+3) on natural indices starting at zero, with d=3≥1. With ε=4/1>0 and candidate N=2, does every n≥N have |u(n)−L|<ε, including n=N?

View answer and explanationYes

Check |c|q<p(N+d) exactly, giving Yes. Equality is not strict enough. The all-tail reduction uses this model's nonincreasing error, not sampling as a proof for an arbitrary sequence.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0069/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-cutoff-validity · candidate





## 70. Strict Eligibility of a Candidate Tail Cutoff

For an arbitrary fixed real L, u(n)=L+(-20)/(n+4) on natural indices starting at zero, with d=4≥1. With ε=5/2>0 and candidate N=4, does every n≥N have |u(n)−L|<ε, including n=N?

View answer and explanationNo

Check |c|q<p(N+d) exactly, giving No. Equality is not strict enough. The all-tail reduction uses this model's nonincreasing error, not sampling as a proof for an arbitrary sequence.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0070/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-cutoff-validity · candidate





## 71. Strict Eligibility of a Candidate Tail Cutoff

For an arbitrary fixed real L, u(n)=L+(21)/(n+1) on natural indices starting at zero, with d=1≥1. With ε=1/3>0 and candidate N=65, does every n≥N have |u(n)−L|<ε, including n=N?

View answer and explanationYes

Check |c|q<p(N+d) exactly, giving Yes. Equality is not strict enough. The all-tail reduction uses this model's nonincreasing error, not sampling as a proof for an arbitrary sequence.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0071/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-cutoff-validity · candidate





## 72. Strict Eligibility of a Candidate Tail Cutoff

For an arbitrary fixed real L, u(n)=L+(-22)/(n+2) on natural indices starting at zero, with d=2≥1. With ε=2/1>0 and candidate N=10, does every n≥N have |u(n)−L|<ε, including n=N?

View answer and explanationYes

Check |c|q<p(N+d) exactly, giving Yes. Equality is not strict enough. The all-tail reduction uses this model's nonincreasing error, not sampling as a proof for an arbitrary sequence.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0072/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-cutoff-validity · candidate





## 73. Strict Eligibility of a Candidate Tail Cutoff

For an arbitrary fixed real L, u(n)=L+(23)/(n+3) on natural indices starting at zero, with d=3≥1. With ε=3/2>0 and candidate N=12, does every n≥N have |u(n)−L|<ε, including n=N?

View answer and explanationNo

Check |c|q<p(N+d) exactly, giving No. Equality is not strict enough. The all-tail reduction uses this model's nonincreasing error, not sampling as a proof for an arbitrary sequence.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0073/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-cutoff-validity · candidate





## 74. Strict Eligibility of a Candidate Tail Cutoff

For an arbitrary fixed real L, u(n)=L+(-24)/(n+4) on natural indices starting at zero, with d=4≥1. With ε=4/3>0 and candidate N=17, does every n≥N have |u(n)−L|<ε, including n=N?

View answer and explanationYes

Check |c|q<p(N+d) exactly, giving Yes. Equality is not strict enough. The all-tail reduction uses this model's nonincreasing error, not sampling as a proof for an arbitrary sequence.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0074/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-cutoff-validity · candidate





## 75. Strict Eligibility of a Candidate Tail Cutoff

For an arbitrary fixed real L, u(n)=L+(25)/(n+1) on natural indices starting at zero, with d=1≥1. With ε=5/1>0 and candidate N=5, does every n≥N have |u(n)−L|<ε, including n=N?

View answer and explanationYes

Check |c|q<p(N+d) exactly, giving Yes. Equality is not strict enough. The all-tail reduction uses this model's nonincreasing error, not sampling as a proof for an arbitrary sequence.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0075/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-cutoff-validity · candidate





## 76. Exact Error at a Specified Sequence Index

For an arbitrary fixed real L, u(n)=L+(0)/(n+1) on natural indices starting at zero, with d=1≥1. Find the exact absolute error |u(0)−L|. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation0/1

Use |c|/(n+d) to obtain 0/1. Retain nonnegative zero-based indexing and the positive denominator; changing L does not affect this error.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0076/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-term-error · candidate





## 77. Exact Error at a Specified Sequence Index

For an arbitrary fixed real L, u(n)=L+(-2)/(n+2) on natural indices starting at zero, with d=2≥1. Find the exact absolute error |u(1)−L|. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation2/3

Use |c|/(n+d) to obtain 2/3. Retain nonnegative zero-based indexing and the positive denominator; changing L does not affect this error.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0077/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-term-error · candidate





## 78. Exact Error at a Specified Sequence Index

For an arbitrary fixed real L, u(n)=L+(3)/(n+3) on natural indices starting at zero, with d=3≥1. Find the exact absolute error |u(2)−L|. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation3/5

Use |c|/(n+d) to obtain 3/5. Retain nonnegative zero-based indexing and the positive denominator; changing L does not affect this error.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0078/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-term-error · candidate





## 79. Exact Error at a Specified Sequence Index

For an arbitrary fixed real L, u(n)=L+(-4)/(n+4) on natural indices starting at zero, with d=4≥1. Find the exact absolute error |u(3)−L|. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation4/7

Use |c|/(n+d) to obtain 4/7. Retain nonnegative zero-based indexing and the positive denominator; changing L does not affect this error.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0079/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-term-error · candidate





## 80. Exact Error at a Specified Sequence Index

For an arbitrary fixed real L, u(n)=L+(5)/(n+1) on natural indices starting at zero, with d=1≥1. Find the exact absolute error |u(4)−L|. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation1/1

Use |c|/(n+d) to obtain 1/1. Retain nonnegative zero-based indexing and the positive denominator; changing L does not affect this error.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0080/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-term-error · candidate





## 81. Exact Error at a Specified Sequence Index

For an arbitrary fixed real L, u(n)=L+(-6)/(n+2) on natural indices starting at zero, with d=2≥1. Find the exact absolute error |u(5)−L|. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation6/7

Use |c|/(n+d) to obtain 6/7. Retain nonnegative zero-based indexing and the positive denominator; changing L does not affect this error.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0081/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-term-error · candidate





## 82. Exact Error at a Specified Sequence Index

For an arbitrary fixed real L, u(n)=L+(7)/(n+3) on natural indices starting at zero, with d=3≥1. Find the exact absolute error |u(6)−L|. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation7/9

Use |c|/(n+d) to obtain 7/9. Retain nonnegative zero-based indexing and the positive denominator; changing L does not affect this error.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0082/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-term-error · candidate





## 83. Exact Error at a Specified Sequence Index

For an arbitrary fixed real L, u(n)=L+(-8)/(n+4) on natural indices starting at zero, with d=4≥1. Find the exact absolute error |u(0)−L|. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation2/1

Use |c|/(n+d) to obtain 2/1. Retain nonnegative zero-based indexing and the positive denominator; changing L does not affect this error.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0083/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-term-error · candidate





## 84. Exact Error at a Specified Sequence Index

For an arbitrary fixed real L, u(n)=L+(9)/(n+1) on natural indices starting at zero, with d=1≥1. Find the exact absolute error |u(1)−L|. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation9/2

Use |c|/(n+d) to obtain 9/2. Retain nonnegative zero-based indexing and the positive denominator; changing L does not affect this error.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0084/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-term-error · candidate





## 85. Exact Error at a Specified Sequence Index

For an arbitrary fixed real L, u(n)=L+(-10)/(n+2) on natural indices starting at zero, with d=2≥1. Find the exact absolute error |u(2)−L|. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation5/2

Use |c|/(n+d) to obtain 5/2. Retain nonnegative zero-based indexing and the positive denominator; changing L does not affect this error.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0085/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-term-error · candidate





## 86. Exact Error at a Specified Sequence Index

For an arbitrary fixed real L, u(n)=L+(11)/(n+3) on natural indices starting at zero, with d=3≥1. Find the exact absolute error |u(3)−L|. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation11/6

Use |c|/(n+d) to obtain 11/6. Retain nonnegative zero-based indexing and the positive denominator; changing L does not affect this error.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0086/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-term-error · candidate





## 87. Exact Error at a Specified Sequence Index

For an arbitrary fixed real L, u(n)=L+(-12)/(n+4) on natural indices starting at zero, with d=4≥1. Find the exact absolute error |u(4)−L|. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation3/2

Use |c|/(n+d) to obtain 3/2. Retain nonnegative zero-based indexing and the positive denominator; changing L does not affect this error.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0087/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-term-error · candidate





## 88. Exact Error at a Specified Sequence Index

For an arbitrary fixed real L, u(n)=L+(13)/(n+1) on natural indices starting at zero, with d=1≥1. Find the exact absolute error |u(5)−L|. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation13/6

Use |c|/(n+d) to obtain 13/6. Retain nonnegative zero-based indexing and the positive denominator; changing L does not affect this error.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0088/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-term-error · candidate





## 89. Exact Error at a Specified Sequence Index

For an arbitrary fixed real L, u(n)=L+(-14)/(n+2) on natural indices starting at zero, with d=2≥1. Find the exact absolute error |u(6)−L|. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation7/4

Use |c|/(n+d) to obtain 7/4. Retain nonnegative zero-based indexing and the positive denominator; changing L does not affect this error.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0089/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-term-error · candidate





## 90. Exact Error at a Specified Sequence Index

For an arbitrary fixed real L, u(n)=L+(15)/(n+3) on natural indices starting at zero, with d=3≥1. Find the exact absolute error |u(0)−L|. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation5/1

Use |c|/(n+d) to obtain 5/1. Retain nonnegative zero-based indexing and the positive denominator; changing L does not affect this error.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0090/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-term-error · candidate





## 91. Exact Error at a Specified Sequence Index

For an arbitrary fixed real L, u(n)=L+(-16)/(n+4) on natural indices starting at zero, with d=4≥1. Find the exact absolute error |u(1)−L|. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation16/5

Use |c|/(n+d) to obtain 16/5. Retain nonnegative zero-based indexing and the positive denominator; changing L does not affect this error.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0091/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-term-error · candidate





## 92. Exact Error at a Specified Sequence Index

For an arbitrary fixed real L, u(n)=L+(17)/(n+1) on natural indices starting at zero, with d=1≥1. Find the exact absolute error |u(2)−L|. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation17/3

Use |c|/(n+d) to obtain 17/3. Retain nonnegative zero-based indexing and the positive denominator; changing L does not affect this error.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0092/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-term-error · candidate





## 93. Exact Error at a Specified Sequence Index

For an arbitrary fixed real L, u(n)=L+(-18)/(n+2) on natural indices starting at zero, with d=2≥1. Find the exact absolute error |u(3)−L|. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation18/5

Use |c|/(n+d) to obtain 18/5. Retain nonnegative zero-based indexing and the positive denominator; changing L does not affect this error.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0093/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-term-error · candidate





## 94. Exact Error at a Specified Sequence Index

For an arbitrary fixed real L, u(n)=L+(19)/(n+3) on natural indices starting at zero, with d=3≥1. Find the exact absolute error |u(4)−L|. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation19/7

Use |c|/(n+d) to obtain 19/7. Retain nonnegative zero-based indexing and the positive denominator; changing L does not affect this error.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0094/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-term-error · candidate





## 95. Exact Error at a Specified Sequence Index

For an arbitrary fixed real L, u(n)=L+(-20)/(n+4) on natural indices starting at zero, with d=4≥1. Find the exact absolute error |u(5)−L|. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation20/9

Use |c|/(n+d) to obtain 20/9. Retain nonnegative zero-based indexing and the positive denominator; changing L does not affect this error.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0095/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-term-error · candidate





## 96. Exact Error at a Specified Sequence Index

For an arbitrary fixed real L, u(n)=L+(21)/(n+1) on natural indices starting at zero, with d=1≥1. Find the exact absolute error |u(6)−L|. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation3/1

Use |c|/(n+d) to obtain 3/1. Retain nonnegative zero-based indexing and the positive denominator; changing L does not affect this error.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0096/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-term-error · candidate





## 97. Exact Error at a Specified Sequence Index

For an arbitrary fixed real L, u(n)=L+(-22)/(n+2) on natural indices starting at zero, with d=2≥1. Find the exact absolute error |u(0)−L|. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation11/1

Use |c|/(n+d) to obtain 11/1. Retain nonnegative zero-based indexing and the positive denominator; changing L does not affect this error.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0097/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-term-error · candidate





## 98. Exact Error at a Specified Sequence Index

For an arbitrary fixed real L, u(n)=L+(23)/(n+3) on natural indices starting at zero, with d=3≥1. Find the exact absolute error |u(1)−L|. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation23/4

Use |c|/(n+d) to obtain 23/4. Retain nonnegative zero-based indexing and the positive denominator; changing L does not affect this error.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0098/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-term-error · candidate





## 99. Exact Error at a Specified Sequence Index

For an arbitrary fixed real L, u(n)=L+(-24)/(n+4) on natural indices starting at zero, with d=4≥1. Find the exact absolute error |u(2)−L|. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation4/1

Use |c|/(n+d) to obtain 4/1. Retain nonnegative zero-based indexing and the positive denominator; changing L does not affect this error.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0099/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-term-error · candidate





## 100. Exact Error at a Specified Sequence Index

For an arbitrary fixed real L, u(n)=L+(25)/(n+1) on natural indices starting at zero, with d=1≥1. Find the exact absolute error |u(3)−L|. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation25/4

Use |c|/(n+d) to obtain 25/4. Retain nonnegative zero-based indexing and the positive denominator; changing L does not affect this error.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r01-sequence-limit-0100/)

Sequence limits and exact tail cutoffs · qf-sequence-limit-term-error · candidate



[Canonical question JSONL](https://ocme.evemisslab.com/data/questions/batches/batch-20261007-advance-20261008-r01-sequence-limit-foundations/questions.jsonl) · [English JSONL](https://ocme.evemisslab.com/data/i18n/questions/batch-20261007-advance-20261008-r01-sequence-limit-foundations.en.jsonl) · [Validation report](https://ocme.evemisslab.com/data/questions/batches/batch-20261007-advance-20261008-r01-sequence-limit-foundations/validation-report.json)