To solve the given limit:

$\lim_{x \to 1} \frac{\ln x}{x^2 - 1}$

Step-by-step solution:

1. Substitution: Substitute $x = 1$ into the expression: $\frac{\ln x}{x^2 - 1} = \frac{\ln(1)}{1^2 - 1} = \frac{0}{0}.$ This results in an indeterminate form ($0/0$), so we apply L'Hôpital's Rule.

2. Apply L'Hôpital's Rule: Differentiate the numerator and denominator with respect to $x$:

   • Derivative of the numerator ($\ln x$) is $\frac{1}{x}$,
   • Derivative of the denominator ($x^2 - 1$) is $2x$.

   Thus, the new expression becomes: $\lim_{x \to 1} \frac{\ln x}{x^2 - 1} = \lim_{x \to 1} \frac{\frac{1}{x}}{2x}.$

3. Simplify: Simplify the fraction: $\frac{\frac{1}{x}}{2x} = \frac{1}{2x^2}.$

4. Evaluate the limit: Substitute $x = 1$ into the simplified expression: $\frac{1}{2(1)^2} = \frac{1}{2}.$

Final Answer:

$\lim_{x \to 1} \frac{\ln x}{x^2 - 1} = \frac{1}{2}.$

Let me know if you'd like further clarification or more examples!

Related Questions:

1. How does L'Hôpital's Rule apply to limits with indeterminate forms?
2. What other methods could solve limits like this without L'Hôpital's Rule?
3. How does the logarithmic function influence the numerator in this limit problem?
4. Can you explain the behavior of the function near $x = 1$?
5. What is the importance of simplifying the derivative step properly when using L'Hôpital's Rule?

Tip:

Always verify that the original limit is indeterminate ($0/0$ or $\infty/\infty$) before applying L'Hôpital's Rule.