To evaluate the limit $\lim_{x \to 0} x \cos(3/x)$ using the Squeeze Theorem, we proceed as follows:

1. Understanding the Function: The function of interest is $x \cos(3/x)$.

2. Behavior of $\cos(3/x)$: We know that $\cos(3/x)$ oscillates between -1 and 1 for all values of $x$ except $x = 0$.

3. Applying Bounds: $-1 \leq \cos(3/x) \leq 1$ Multiplying each part of the inequality by $x$ (and noting the change of signs if $x$ is negative), we obtain: $-x \leq x \cos(3/x) \leq x$

4. Limits of the Bounding Functions:

   • As $x \to 0$, both $x$ and $-x$ approach 0.

5. Applying the Squeeze Theorem: Since $-x \leq x \cos(3/x) \leq x$ and both $-x$ and $x$ approach 0 as $x \to 0$, it follows by the Squeeze Theorem that: $\lim_{x \to 0} x \cos(3/x) = 0$

Hence, the limit of $x \cos(3/x)$ as $x \to 0$ is 0.

Would you like more details on any part of this explanation or have any other questions?

Here are some follow-up topics you might find interesting:

1. What would change if the limit $x \to a$ for some $a \neq 0$?
2. How does the Squeeze Theorem apply to limits involving trigonometric functions generally?
3. Could we use another method to solve this limit?
4. What happens to the limit if the function inside the cosine is different, like $\cos(5/x)$?
5. How do we rigorously prove that the function $\cos(3/x)$ stays within -1 and 1?

Tip: When applying the Squeeze Theorem, it's crucial to ensure the function in question is indeed "squeezed" by the bounding functions at the limit point.