To evaluate the limit:

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

we can proceed by analyzing the behavior of the function as $x \to 0$.

Step-by-Step Solution

1. Identify the Form: As $x \to 0$, both $e^x - 1$ and $\tan x$ approach 0, giving us an indeterminate form $\frac{0}{0}$. Therefore, we can apply L'Hôpital's Rule, which states that for limits of the form $\frac{0}{0}$, we can take the derivatives of the numerator and denominator.

2. Differentiate the Numerator and Denominator:

   • The derivative of the numerator $e^x - 1$ is $e^x$.
   • The derivative of the denominator $\tan x$ is $\sec^2 x$.

   So the limit becomes: $\lim_{x \to 0} \frac{e^x}{\sec^2 x} = \lim_{x \to 0} e^x \cos^2 x$

3. Evaluate the Limit: Now, substitute $x = 0$ into $e^x \cos^2 x$.

   • $e^0 = 1$
   • $\cos^2 0 = 1$

   Thus: $\lim_{x \to 0} e^x \cos^2 x = 1 \cdot 1 = 1$

Final Answer

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

Would you like further details or have any questions?

Here are some related questions that might interest you:

1. How would the limit change if we had $\lim_{x \to 0} \frac{\ln(1+x)}{\tan x}$?
2. What if we replaced $e^x - 1$ with $\sin x$ in this limit?
3. Can we use Taylor expansions to approximate limits like this one?
4. How does the behavior of $e^x - 1$ compare to $\tan x$ as $x \to 0$?
5. What is L'Hôpital's Rule, and when should it be applied?

Tip: In problems involving indeterminate forms, L'Hôpital's Rule is a powerful tool, but always remember to check if simpler methods (like factoring or simplification) can solve the problem more efficiently.