To find the limit of \(\lim_{x \to 0} \frac{e^{3x} - 1}{x}\), we can use L'Hôpital's Rule. L'Hôpital's Rule applies to limits of the form \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\).

First, let's check the form of the limit:

As \(x \to 0\):
- \(e^{3x} \to e^0 = 1\)
- \(e^{3x} - 1 \to 0\)
- \(x \to 0\)

So, the limit is of the indeterminate form \(\frac{0}{0}\). Therefore, we can apply L'Hôpital's Rule, which states:

\[
\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}
\]

if the limit on the right exists. Here, \(f(x) = e^{3x} - 1\) and \(g(x) = x\).

We find the derivatives:
- \(f'(x) = \frac{d}{dx}(e^{3x} - 1) = 3e^{3x}\)
- \(g'(x) = \frac{d}{dx}(x) = 1\)

Now we apply L'Hôpital's Rule:

\[
\lim_{x \to 0} \frac{e^{3x} - 1}{x} = \lim_{x \to 0} \frac{3e^{3x}}{1}
\]

As \(x \to 0\), \(e^{3x} \to e^0 = 1\), so:

\[
\lim_{x \to 0} \frac{3e^{3x}}{1} = 3 \cdot 1 = 3
\]

Therefore, the limit is:

\[
\boxed{3}
\]

Would you like any further details or have any questions about this process? Here are some related questions:

1. What is L'Hôpital's Rule and when can it be applied?
2. How do you find the derivative of an exponential function?
3. Why does \(e^{3x} \to 1\) as \(x \to 0\)?
4. What are indeterminate forms, and why are they important in calculus?
5. How can you verify the result of a limit problem using numerical approximation?
6. What is the significance of exponential functions in calculus?
7. How do exponential functions behave as their input approaches infinity or negative infinity?
8. Can L'Hôpital's Rule be applied multiple times in solving a limit?

**Tip:** Always check if a limit problem results in an indeterminate form before applying L'Hôpital's Rule. This ensures the rule is used correctly.

Learn how to solve the limit problem \( \lim_{x \to 0} \frac{e^{3x} - 1}{x} \) using L'Hôpital's Rule. This problem involves exponential functions, derivatives, and indeterminate forms. Suitable for advanced calculus students.

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