To prove that 

\[
\lim_{x \to 0} x \sin\left(\frac{1}{x}\right) = 0,
\]

we can use the Squeeze Theorem (also known as the Sandwich Theorem).

### Step-by-Step Proof

1. **Understand the Function Behavior:**

   The sine function, \(\sin\left(\frac{1}{x}\right)\), oscillates between \(-1\) and \(1\) for all values of \(x\), so we have:

   \[
   -1 \leq \sin\left(\frac{1}{x}\right) \leq 1.
   \]

2. **Multiply by \(x\):**

   Multiply the entire inequality by \(x\). Since \(x\) can be positive or negative, we'll consider two cases:

   - For \(x > 0\):

     \[
     -x \leq x \sin\left(\frac{1}{x}\right) \leq x.
     \]

   - For \(x < 0\):

     \[
     x \leq x \sin\left(\frac{1}{x}\right) \leq -x.
     \]

   Notice that these inequalities hold for \(x\) approaching 0 from either side.

3. **Consider the Limit of the Bounds:**

   As \(x \to 0\), both \(x\) and \(-x\) approach 0. Therefore:

   \[
   \lim_{x \to 0^+} x = 0 \quad \text{and} \quad \lim_{x \to 0^+} -x = 0.
   \]

   Similarly,

   \[
   \lim_{x \to 0^-} x = 0 \quad \text{and} \quad \lim_{x \to 0^-} -x = 0.
   \]

   Hence, in either case:

   \[
   \lim_{x \to 0} x \sin\left(\frac{1}{x}\right) = 0.
   \]

4. **Apply the Squeeze Theorem:**

   Since \(x \sin\left(\frac{1}{x}\right)\) is squeezed between two functions that both tend to 0 as \(x \to 0\), by the Squeeze Theorem:

   \[
   \lim_{x \to 0} x \sin\left(\frac{1}{x}\right) = 0.
   \]

This completes the proof.

Would you like any further details or explanations on this proof?

Here are 5 related questions to expand your understanding:

1. How does the Squeeze Theorem work in different types of limits?
2. Can you prove the limit using an epsilon-delta definition?
3. What happens to the function \(x \sin\left(\frac{1}{x}\right)\) as \(x\) approaches infinity?
4. How does the limit change if we replace \(x \sin\left(\frac{1}{x}\right)\) with \(x^2 \sin\left(\frac{1}{x}\right)\)?
5. What are some other common functions that require the Squeeze Theorem for limit proofs?

**Tip:** When working with limits involving oscillating functions, the Squeeze Theorem is a powerful tool, especially if you can bound the function between two simpler functions that have known limits.

Learn how to prove the limit of x sin(1/x) equals 0 using the Squeeze Theorem with a detailed step-by-step solution. This problem involves understanding limits, trigonometric functions, and applying the Squeeze Theorem. Suitable for undergraduate students and those interested in advanced calculus proofs.

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