The expression in the image is a limit:

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

This is a standard limit in calculus. To evaluate it, we can recognize that this is a definition for the derivative of $e^x$ at $x = 0$.

We can also solve it by applying L'Hopital's Rule, since both the numerator and denominator tend to 0 as $x \to 0$.

Applying L'Hopital's Rule:

1. Take the derivative of the numerator $e^x - 1$, which is $e^x$.
2. Take the derivative of the denominator $x$, which is 1.

Thus, the limit becomes:

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

Therefore, the value of the limit is 1.

Would you like further details or clarification on this? Let me know if you have any questions.

Here are 5 related questions:

1. What is the definition of the derivative of a function?
2. How do you apply L'Hopital's Rule to limits?
3. Why does $e^x$ have the property that its derivative is itself?
4. What is the limit of $\frac{\sin(x)}{x}$ as $x \to 0$?
5. How do you determine if a limit results in an indeterminate form like $\frac{0}{0}$?

Tip: For limits involving indeterminate forms, L'Hopital's Rule is often a helpful tool.