The problem shown in the image is:

$\lim_{x \to 0} \frac{e^x - e^{-x}}{x}$

Solution:

1. Substitute $x = 0$: Direct substitution gives: $\frac{e^0 - e^0}{0} = \frac{0}{0}$ This is an indeterminate form, so we apply L'Hôpital's Rule.

2. Apply L'Hôpital's Rule: According to L'Hôpital's Rule, for limits in indeterminate forms like $\frac{0}{0}$: $\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}$ Differentiate the numerator and denominator:

   • Numerator: $f(x) = e^x - e^{-x} \implies f'(x) = e^x + e^{-x}$
   • Denominator: $g(x) = x \implies g'(x) = 1$

   Substitute into the limit: $\lim_{x \to 0} \frac{e^x - e^{-x}}{x} = \lim_{x \to 0} \frac{e^x + e^{-x}}{1}$

3. Evaluate the new limit: Substitute $x = 0$: $e^x + e^{-x} = e^0 + e^0 = 1 + 1 = 2$

   Thus: $\lim_{x \to 0} \frac{e^x - e^{-x}}{x} = 2$

Final Answer:

$\boxed{2}$

Would you like a detailed explanation of the steps, or do you have any related questions? Here are 5 potential follow-up questions:

1. What is L'Hôpital's Rule, and when can it be used?
2. How do you differentiate exponential functions like $e^x$ and $e^{-x}$?
3. What are some other examples of indeterminate forms in limits?
4. Can this problem be solved without L'Hôpital's Rule? If so, how?
5. What is the geometric interpretation of this limit?

Tip: Always check if substitution works directly before applying advanced techniques like L'Hôpital's Rule.