Limit of \( \displaystyle \frac{\tan{\left(x - 1 \right)}}{x - 1} \) as \( x \to 1 \)
Problem 1.54 · easy
Evaluate \( \displaystyle \lim_{x \to 1} \frac{\tan{\left(x - 1 \right)}}{x - 1} \).
- \[ \lim_{x \to 1^+}\left(\frac{\tan{\left(x - 1 \right)}}{x - 1}\right) \]limitStart with the original limit.✓ Proved
- \[ = \lim_{x \to 1^+}\left(\frac{\sin{\left(x - 1 \right)}}{\left(x - 1\right) \cos{\left(x - 1 \right)}}\right) \]rewrite algebraUse the identity tan(u) = sin(u)/cos(u). Rearrange the fraction.✓ Proved
- \[ = \lim_{x \to 1^+}\left(\frac{\sin{\left(x - 1 \right)}}{x - 1}\right) \lim_{x \to 1^+} \frac{1}{\cos{\left(x - 1 \right)}} \]limit-lawApply the product rule for limits.✓ Proved
- \[ = \lim_{x \to 1^+}\left(\frac{\frac{d}{d x} \sin{\left(x - 1 \right)}}{\frac{d}{d x} \left(x - 1\right)}\right) \lim_{x \to 1^+} \frac{1}{\cos{\left(x - 1 \right)}} \]lhopitalApply L'Hopital's rule to the first limit.✓ Proved
- \[ = \lim_{x \to 1^+} \cos{\left(x - 1 \right)} \lim_{x \to 1^+} \frac{1}{\cos{\left(x - 1 \right)}} \]simplifyEvaluate the derivatives.✓ Proved
- \[ = \lim_{x \to 1^+} 1 \]algebra simplifyCombine the terms. Simplify the expression inside the limit.✓ Proved
- \[ = 1 \]limitEvaluate the final limit.✓ Proved
Answer \( 1 \)
✓ Nihil obstat Every line of this solution was proved by the computer algebra system SymPy. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation.
The full receipt
| Line | Status | Checked by | Detail |
|---|---|---|---|
| 1 | ✓ Proved | sympy 1.14.0 | line 1 is the problem as stated |
| 2 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 tan has poles at odd multiples of pi/2 undefined where x - 1 = 0 undefined where cos(x - 1) = 0 |
| 3 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 undefined where cos(x - 1) = 0 undefined where x - 1 = 0 |
| 4 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 undefined where cos(x - 1) = 0 undefined where x - 1 = 0 |
| 5 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 undefined where cos(x - 1) = 0 undefined where x - 1 = 0 undefined where Derivative(x - 1, x) = 0 |
| 6 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 undefined where cos(x - 1) = 0 undefined where Derivative(x - 1, x) = 0 |
| 7 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 undefined where cos(x - 1) = 0 |
| 8 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 |
| 9 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 |
| answer | ✓ Proved | sympy 1.14.0 | final line against the stated answer: simplify(a - b) reduced to 0 |
| answer, a second way | ✓ Proved | sympy 1.14.0 | SymPy took the limit from both sides and got the stated value |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: pass — The solution correctly applies L'Hopital's rule to the indeterminate form and uses valid limit laws. Each step changes only one aspect of the expression and uses labels from the allowed vocabulary.
Every verdict on record (4)
qwen3.6:27b-mlx: pass 2026-10-03 — The solution correctly applies L'Hopital's rule to the indeterminate form and uses valid limit laws. Each step changes only one aspect of the expression and uses labels from the allowed vocabulary.gpt-oss:20b: pass 2026-10-03qwen3.6:27b-mlx: pass 2026-09-30 — The solution correctly applies L'Hopital's rule to the indeterminate form and simplifies the resulting expression. Each step changes only one aspect of the expression and uses valid labels from the fixed vocabulary.gpt-oss:20b: pass 2026-09-30
Proved: SymPy reduced the difference between the two sides to zero. Checked independently: a separate
method, named above, confirmed it. Checked numerically: the two sides agree at every sampled point, which is evidence,
not proof. Reviewed: a model or a person read it; that is all a sentence can have. Solution by
gemma4:26b, checked 2026-10-03 with SymPy 1.14.0.