Let's analyze and solve each problem from the image you uploaded:

Problem 1:

We need to show that for all nonzero $x \in \mathbb{R}$: $-|x| \leq x \sin\left(\frac{1}{x}\right) \leq |x|.$

Solution:

1. Bound the sine function: Recall that for any real number $y$, the sine function satisfies: $-1 \leq \sin(y) \leq 1.$ In this case, we are dealing with $\sin\left(\frac{1}{x}\right)$, so we have: $-1 \leq \sin\left(\frac{1}{x}\right) \leq 1.$

2. Multiply by $x$: We multiply the entire inequality by $x$. However, we must be careful about the sign of $x$. Let's analyze two cases:

   • If $x > 0$, multiplying by a positive $x$ preserves the inequality: $-x \leq x \sin\left(\frac{1}{x}\right) \leq x.$

   • If $x < 0$, multiplying by a negative $x$ reverses the inequality: $x \leq x \sin\left(\frac{1}{x}\right) \leq -x.$ In both cases, this inequality can be rewritten as: $-|x| \leq x \sin\left(\frac{1}{x}\right) \leq |x|.$ Hence, we have shown the desired inequality.

Problem 2:

We are asked to compute the following limit: $\lim_{x \to 0} x \sin\left(\frac{1}{x}\right).$

Solution:

1. Bound the function: From Problem 1, we already have the inequality: $-|x| \leq x \sin\left(\frac{1}{x}\right) \leq |x|.$

2. Apply the Squeeze Theorem: Since $-|x| \leq x \sin\left(\frac{1}{x}\right) \leq |x|$ and we know that $\lim_{x \to 0} |x| = 0$, by the Squeeze Theorem, we conclude: $\lim_{x \to 0} x \sin\left(\frac{1}{x}\right) = 0.$

Thus, the limit is 0.

Would you like a more detailed explanation of any of these steps or have any additional questions? Here are some related questions to explore further:

1. How does the Squeeze Theorem work, and when can it be applied?
2. Why is the inequality $-|x| \leq x \sin\left(\frac{1}{x}\right) \leq |x|$ important in solving limits?
3. Can the Squeeze Theorem be applied to other trigonometric functions? If so, how?
4. What would happen to the limit if $\sin\left(\frac{1}{x}\right)$ were replaced with $\cos\left(\frac{1}{x}\right)$?
5. How does the behavior of $x \sin\left(\frac{1}{x}\right)$ change as $x$ approaches other values (e.g., $x \to \infty$)?

Tip: The Squeeze Theorem is a powerful tool in calculus for dealing with limits involving oscillatory functions like sine and cosine.