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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.

Explore how to solve the limit of (e^x - 1)/x using L'Hôpital's Rule with a step-by-step explanation. Understand the conditions for applying L'Hôpital's Rule and its application in calculus. Suitable for advanced high school and college students.

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