To solve the given limit:

$\lim_{x \to 0} \left( e^{2x} - 1 \right) \cdot \cot(3x)$

Step 1: Simplify using approximations for small $x$

For small $x$, the following approximations can be applied:

1. $e^{2x} - 1 \approx 2x$, as the Taylor expansion of $e^{2x}$ is $1 + 2x + \frac{(2x)^2}{2!} + \dots$, so $e^{2x} - 1 \approx 2x$.
2. $\cot(3x) = \frac{\cos(3x)}{\sin(3x)} \approx \frac{1}{3x}$, because $\sin(3x) \approx 3x$ and $\cos(3x) \to 1$ as $x \to 0$.

Step 2: Substitute approximations

$\lim_{x \to 0} \left( e^{2x} - 1 \right) \cdot \cot(3x) \approx \lim_{x \to 0} (2x) \cdot \frac{1}{3x}.$

Step 3: Simplify

$\lim_{x \to 0} \frac{2x}{3x} = \lim_{x \to 0} \frac{2}{3} = \frac{2}{3}.$

Final Answer:

$\boxed{\frac{2}{3}}$

Let me know if you would like further explanation or a deeper breakdown of the steps!

Relative Questions:

1. What is the Taylor expansion of $e^x$ and how does it apply here?
2. Why can $\sin(x) \approx x$ and $\cos(x) \approx 1$ be used for small values of $x$?
3. How is the approximation for $\cot(x)$ derived?
4. What happens if higher-order terms are included in the expansion?
5. Can we approach this problem without approximations using L'Hôpital's Rule?

Tip:

For small $x$, approximations like $\sin(x) \approx x$ and $e^x \approx 1 + x$ are extremely useful to simplify limits quickly!