The image provided contains Example 1, which appears to be focused on solving limits.

Let's walk through the detailed solution process step by step:

Problem

We are asked to find the limit:

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

Step-by-step Solution

1. Recognize the Standard Limit Form: The expression resembles a well-known standard limit form: $\lim_{{x \to 0}} \frac{e^x - 1}{x} = 1$ This result comes from the derivative of the exponential function at 0. Specifically, the derivative of $e^x$ at $x = 0$ is 1. So we can directly apply this standard limit formula.

2. Application of L'Hopital's Rule (if necessary): If you were not familiar with the standard form, you could solve it using L'Hopital's Rule. L'Hopital's Rule applies to limits of indeterminate forms like $\frac{0}{0}$. To apply the rule, take the derivatives of the numerator and denominator:

   • The derivative of $e^x - 1$ is $e^x$
   • The derivative of $x$ is 1.

   So applying L'Hopital's Rule gives us: $\lim_{{x \to 0}} \frac{e^x - 1}{x} = \lim_{{x \to 0}} \frac{e^x}{1} = e^0 = 1$

3. Conclusion: The limit evaluates to $1$, either by applying the standard result or by using L'Hopital's Rule.

$\boxed{1}$

This is the detailed explanation of the example in your image. If you'd like to delve further into limits or related topics, feel free to ask!

Additional Relative Questions:

1. What are other standard limit forms similar to the one used here?
2. How does L'Hopital's Rule apply in more complicated limit problems?
3. Can the limit $\lim_{{x \to 0}} \frac{e^x - 1}{x}$ be solved using a series expansion of $e^x$?
4. How do limits involving trigonometric functions differ from exponential functions?
5. What role does continuity play in determining limits of functions like $e^x$?

Tip: When dealing with limits, knowing key standard forms like $\lim_{{x \to 0}} \frac{\sin x}{x} = 1$ and $\lim_{{x \to 0}} \frac{e^x - 1}{x} = 1$ can save you time in computations.