To evaluate the limit

$\lim_{x \to 1} \frac{\ln x}{14x - x^2 - 13},$

we first substitute $x = 1$:

$\ln(1) = 0,$ $14(1) - (1)^2 - 13 = 14 - 1 - 13 = 0.$

This gives us the indeterminate form $\frac{0}{0}$. Thus, we can apply l'Hôpital's Rule.

Step 1: Differentiate the numerator and the denominator

1. Numerator: The derivative of $\ln x$ is $\frac{1}{x}$.
2. Denominator: The derivative of $14x - x^2 - 13$ is $14 - 2x$.

Step 2: Rewrite the limit using l'Hôpital's Rule

Using l'Hôpital's Rule, we rewrite the limit:

$\lim_{x \to 1} \frac{\ln x}{14x - x^2 - 13} = \lim_{x \to 1} \frac{\frac{1}{x}}{14 - 2x}.$

Step 3: Evaluate the new limit

Now we substitute $x = 1$:

$\frac{\frac{1}{1}}{14 - 2(1)} = \frac{1}{14 - 2} = \frac{1}{12}.$

Conclusion

Thus, the limit evaluates to

$\lim_{x \to 1} \frac{\ln x}{14x - x^2 - 13} = \frac{1}{12}.$

If you have any questions or need further details, feel free to ask!

Related Questions:

1. How do you apply l'Hôpital's Rule in different types of indeterminate forms?
2. What are the conditions under which l'Hôpital's Rule can be used?
3. Can you provide examples of limits that do not require l'Hôpital's Rule?
4. What is the significance of continuity in evaluating limits?
5. How does the derivative of a function relate to the concept of limits?

Tip:

Always check if you can simplify the expression before applying l'Hôpital's Rule, as it may lead to an easier evaluation.