Question

The value of \( \lim_{t \to 1} \frac{\ln t}{t^2 - 1} \) is ________.

• A. \( 1 \)
• B. \( 0 \)
• C. \( 1 \)
• D. \( \infty \)

Hint:

When encountering indeterminate forms like \( \frac{0}{0} \), apply L'Hopital's Rule: differentiate the numerator and denominator separately and then take the limit.

Answer 1

The Correct Option is A

Step 1: Substitute \( t = 1 \) into the expression.
We start by substituting \( t = 1 \) into the limit expression: \[ \lim_{t \to 1} \frac{\ln t}{t^2 - 1} = \frac{\ln(1)}{1^2 - 1} = \frac{0}{0}. \] This results in an indeterminate form of \( \frac{0}{0} \), so we need to apply L'Hopital's Rule. 
Step 2: Apply L'Hopital's Rule.
L'Hopital's Rule states that if the limit results in a \( \frac{0}{0} \) indeterminate form, we differentiate the numerator and denominator separately and then take the limit. Let's differentiate the numerator and denominator:
The derivative of \( \ln t \) is \( \frac{1}{t} \).
The derivative of \( t^2 - 1 \) is \( 2t \).
Therefore, applying L'Hopital's Rule, we get: \[ \lim_{t \to 1} \frac{\ln t}{t^2 - 1} = \lim_{t \to 1} \frac{\frac{1}{t}}{2t}. \] Step 3: Simplify and evaluate the limit.
Now simplify the expression: \[ \frac{\frac{1}{t}}{2t} = \frac{1}{2t^2}. \] Substitute \( t = 1 \): \[ \frac{1}{2(1)^2} = \frac{1}{2}. \] Thus, the value of the limit is \( 1 \). 
Step 4: Conclusion.
Therefore, the correct value of the limit is \( 1 \), so the answer is \( \boxed{A} \).