The image contains an expression where the L'Hôpital's rule needs to be applied. I will first transcribe and solve it for you.

Expression:

$\lim_{{x \to 0}} \frac{e^x - 1}{x}$

Solution using L'Hôpital's Rule:

L'Hôpital's Rule can be applied when you have an indeterminate form of the type $\frac{0}{0}$ or $\frac{\infty}{\infty}$. The expression given is indeed of the form $\frac{0}{0}$ as both the numerator $e^x - 1$ and the denominator $x$ approach 0 as $x$ approaches 0.

1. Differentiate the numerator and denominator:

   • Numerator: $\frac{d}{dx}(e^x - 1) = e^x$
   • Denominator: $\frac{d}{dx}(x) = 1$

2. Apply L'Hôpital's Rule: $\lim_{{x \to 0}} \frac{e^x - 1}{x} = \lim_{{x \to 0}} \frac{e^x}{1} = e^0 = 1$

Final Answer:

$\lim_{{x \to 0}} \frac{e^x - 1}{x} = 1$

Would you like further details or explanations on any part of this solution?

Relative Questions:

1. What are the conditions for applying L'Hôpital's Rule?
2. How would the limit change if the expression inside the limit was different, like $\frac{\ln(1+x)}{x}$?
3. What happens to the limit if $x$ approaches infinity instead of zero?
4. Can you use L'Hôpital's Rule for limits involving trigonometric functions?
5. How does L'Hôpital's Rule extend to cases where multiple applications are needed?

Tip:

When applying L'Hôpital's Rule, always check that the conditions for the rule are met (i.e., the form is indeterminate) before proceeding.