To compute the limit $\lim_{x \to 0} x \sin\left(\frac{1}{x}\right)$, let's analyze it step by step.

Step 1: Check for behavior as $x \to 0$

The function involves $\sin\left(\frac{1}{x}\right)$, which oscillates rapidly as $x \to 0$ because $\frac{1}{x} \to \infty$. However, we are multiplying this oscillating function by $x$, which approaches 0.

Step 2: Use the Squeeze Theorem

We can use the fact that $\sin$ is bounded: $-1 \leq \sin\left(\frac{1}{x}\right) \leq 1$ Multiplying the entire inequality by $x$, we get: $-x \leq x \sin\left(\frac{1}{x}\right) \leq x$ As $x \to 0$, both $-x$ and $x$ approach 0.

Step 3: Apply the Squeeze Theorem

Since $-x \leq x \sin\left(\frac{1}{x}\right) \leq x$ and both bounds tend to 0 as $x \to 0$, by the Squeeze Theorem, we conclude: $\lim_{x \to 0} x \sin\left(\frac{1}{x}\right) = 0$

Final Answer:

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

Do you want more details or have any questions? Here are 5 related questions you might find helpful:

1. How does the Squeeze Theorem work in general?
2. What happens to the limit if we replace $x \sin\left(\frac{1}{x}\right)$ with $x^2 \sin\left(\frac{1}{x}\right)$?
3. Can we apply L'Hopital's Rule to this limit?
4. How do we handle limits involving oscillating functions like sine or cosine?
5. What are other methods for computing limits besides the Squeeze Theorem?

Tip: When dealing with oscillating functions multiplied by small values, consider using the Squeeze Theorem to estimate the behavior of the product.