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### 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.

Show that, for all nonzero x ∈ ℝ, -|x| ≤ x sin(1/x) ≤ |x|. Compute lim x→0 x sin(1/x).

Explore the proof of the inequality −|x| ≤ x sin(1/x) ≤ |x| for nonzero real x, and learn how to compute lim_{x→0} x sin(1/x). This involves trigonometry and the Squeeze Theorem, making it ideal for advanced high school or early university students.

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