The image shows the following mathematical limit:

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

Solution:

To evaluate the limit, observe the following:

1. The expression involves $e^x$, the exponential function, which is differentiable and continuous.
2. At $x = 0$, both the numerator $e^x - 1$ and denominator $x$ approach $0$. Hence, it is an indeterminate form $\frac{0}{0}$, so we can apply L'Hôpital's Rule.

Step 1: Differentiating the numerator and denominator

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

Step 2: Apply L'Hôpital's Rule

$\lim_{x \to 0} \frac{e^x - 1}{x} = \lim_{x \to 0} \frac{\frac{d}{dx}(e^x - 1)}{\frac{d}{dx}(x)} = \lim_{x \to 0} \frac{e^x}{1}.$

Step 3: Evaluate the limit

Substitute $x = 0$ into $e^x$: $\lim_{x \to 0} e^x = e^0 = 1.$

Final Answer:

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

Let me know if you'd like a detailed explanation of L'Hôpital's Rule or the exponential function properties.

Related Questions:

1. What is the Taylor series expansion of $e^x$ around $x = 0$?
2. Can L'Hôpital's Rule be used for higher-order indeterminate forms like $0/0$ or $\infty/\infty$?
3. Evaluate $\lim_{x \to 0} \frac{\sin x}{x}$ and explain why it equals 1.
4. Derive the limit of $\frac{\ln(1+x)}{x}$ as $x \to 0$.
5. How do you prove the derivative of $e^x$ using first principles?

Tip:

L'Hôpital's Rule is very useful for handling indeterminate forms $\frac{0}{0}$ or $\frac{\infty}{\infty}$, but it applies only when the derivatives exist and the limit remains indeterminate after differentiation. Always check the form of the limit before applying the rule!