To evaluate the limit of $x \cot(x)$ as $x \to 0$, we proceed as follows:

We know that: $\cot(x) = \frac{\cos(x)}{\sin(x)}$

Thus, the expression $x \cot(x)$ becomes: $x \cot(x) = x \cdot \frac{\cos(x)}{\sin(x)} = \frac{x \cos(x)}{\sin(x)}$

Now, we evaluate the limit: $\lim_{x \to 0} \frac{x \cos(x)}{\sin(x)}$

We can split this into two parts: $\lim_{x \to 0} \frac{x}{\sin(x)} \cdot \lim_{x \to 0} \cos(x)$

From the standard limit, we know: $\lim_{x \to 0} \frac{x}{\sin(x)} = 1$ and $\lim_{x \to 0} \cos(x) = 1$

Therefore: $\lim_{x \to 0} x \cot(x) = 1 \cdot 1 = 1$

So, the value of the limit is: $\boxed{1}$

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Related questions:

1. How do you evaluate limits involving trigonometric functions in general?
2. What is L'Hopital's Rule, and can it be used for this limit?
3. How do small-angle approximations help in limit evaluations?
4. What is the limit $\lim_{x \to 0} \frac{\sin(x)}{x}$, and why is it important?
5. How does the continuity of cosine and sine functions affect their limits?

Tip: Remember that $\lim_{x \to 0} \frac{\sin(x)}{x} = 1$ is a crucial result for evaluating many trigonometric limits.