The limit expression you provided is:

$\lim_{x \to 0} x \sin\left(\frac{1}{x^2}\right)$

Solution:

We can analyze this limit using the Squeeze Theorem (also called the Sandwich Theorem), because the sine function is bounded between -1 and 1. This means:

$-1 \leq \sin\left(\frac{1}{x^2}\right) \leq 1$

Multiplying all sides of the inequality by $x$ (since we are considering the limit as $x \to 0$, and we assume $x > 0$):

$-x \leq x \sin\left(\frac{1}{x^2}\right) \leq x$

Now, as $x \to 0$, both the lower bound $-x$ and the upper bound $x$ tend to 0. So, by the Squeeze Theorem:

$\lim_{x \to 0} x \sin\left(\frac{1}{x^2}\right) = 0$

Final Answer:

$\lim_{x \to 0} x \sin\left(\frac{1}{x^2}\right) = 0$

Would you like more details on the Squeeze Theorem or have any further questions?

Here are 5 related questions to expand on this topic:

1. How does the Squeeze Theorem work in general, and when should it be used?
2. What happens if we try to compute the limit of $\sin\left(\frac{1}{x}\right)$ as $x \to 0$?
3. How does the behavior of trigonometric functions affect limit calculations like this one?
4. Can L'Hôpital's Rule be applied to this type of problem, and why or why not?
5. How would this limit change if the sine function were replaced by another bounded function like $\cos$?

Tip: When applying the Squeeze Theorem, always look for functions that are bounded (like $\sin(x)$ and $\cos(x)$) to help establish the bounds needed for the theorem.