We are asked to evaluate the limit:

$\lim_{x \to 0} \frac{\exp(3x) - 1}{x}$

Step 1: Apply L'Hopital's Rule

The limit is of the indeterminate form $\frac{0}{0}$, as:

• $\exp(3x) \to 1$ as $x \to 0$,
• The numerator $\exp(3x) - 1 \to 0$,
• The denominator $x \to 0$.

L'Hopital's Rule is applicable when the limit results in the form $\frac{0}{0}$. It states that:

$\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}$ if the limit on the right-hand side exists.

Step 2: Differentiate the Numerator and Denominator

We differentiate the numerator and denominator separately.

• The derivative of the numerator $\exp(3x) - 1$ with respect to $x$ is:

$\frac{d}{dx}\left(\exp(3x) - 1\right) = 3\exp(3x)$

• The derivative of the denominator $x$ with respect to $x$ is:

$\frac{d}{dx}(x) = 1$

Step 3: Apply the Rule

Using L'Hopital's Rule:

$\lim_{x \to 0} \frac{\exp(3x) - 1}{x} = \lim_{x \to 0} \frac{3\exp(3x)}{1}$

As $x \to 0$, $\exp(3x) \to 1$, so:

$\lim_{x \to 0} \frac{3\exp(3x)}{1} = 3 \times 1 = 3$

Thus, the value of the limit is:

$\boxed{3}$

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Would you like further details on the solution process or any clarifications? Here are some related questions that could deepen the understanding:

1. What are other indeterminate forms where L'Hopital's Rule can be applied?
2. How does the exponential function behave as $x \to 0$?
3. Can this result be generalized for other exponential expressions such as $\exp(kx)$?
4. How can we derive the limit using a series expansion for $\exp(3x)$?
5. What are the applications of limits involving exponential functions in calculus?

Tip: L'Hopital's Rule can be very useful for indeterminate forms like $\frac{0}{0}$ or $\frac{\infty}{\infty}$. Always check if it applies before trying alternative methods like series expansion.