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The given problem asks you to evaluate the following limit:

\[
\lim_{x \to 0} x \sin \left( \frac{9}{x} \right)
\]

using the Squeeze Theorem. Additionally, it is given that:

\[
-|x| \leq x \sin\left( \frac{9}{x} \right) \leq |x| \quad \text{for all } x \neq 0.
\]

Let's proceed step by step to solve the problem.

### Step 1: Understanding the Bounds
The expression given tells us that the function \(x \sin\left( \frac{9}{x} \right)\) is squeezed between \(-|x|\) and \(|x|\):

\[
-|x| \leq x \sin\left( \frac{9}{x} \right) \leq |x| \quad \text{for all } x \neq 0.
\]

### Step 2: Setting Up the Squeeze Theorem
The Squeeze Theorem states that if we have a function \(f(x)\) that lies between two other functions \(g(x)\) and \(h(x)\), and if:

\[
\lim_{x \to a} g(x) = \lim_{x \to a} h(x) = L,
\]

then:

\[
\lim_{x \to a} f(x) = L.
\]

In our case:
- \(f(x) = x \sin\left( \frac{9}{x} \right)\)
- \(g(x) = -|x|\)
- \(h(x) = |x|\)

We have:

\[
-|x| \leq x \sin\left( \frac{9}{x} \right) \leq |x|
\]

### Step 3: Taking the Limit of the Bound Functions
Let's evaluate the limits of \(g(x)\) and \(h(x)\) as \(x \to 0\):

\[
\lim_{x \to 0} -|x| = 0, \quad \lim_{x \to 0} |x| = 0
\]

Since both \(-|x|\) and \(|x|\) approach \(0\) as \(x \to 0\), we can use the Squeeze Theorem.

### Step 4: Applying the Squeeze Theorem
By the Squeeze Theorem, since:

\[
-|x| \leq x \sin\left( \frac{9}{x} \right) \leq |x| \quad \text{and} \quad \lim_{x \to 0} -|x| = \lim_{x \to 0} |x| = 0,
\]

it follows that:

\[
\lim_{x \to 0} x \sin\left( \frac{9}{x} \right) = 0
\]

### Final Answer
\[
\lim_{x \to 0} x \sin\left( \frac{9}{x} \right) = 0
\]

Would you like me to provide more details or explain any part further?

Here are some related questions to deepen your understanding:

1. How does the Squeeze Theorem help in evaluating limits with oscillating functions?
2. What is the behavior of \(\sin\left( \frac{9}{x} \right)\) as \(x\) approaches 0?
3. Can you apply the Squeeze Theorem to \(\lim_{x \to 0} x \cos\left( \frac{1}{x} \right)\)? If so, what is the result?
4. What are the limitations of using the Squeeze Theorem?
5. How would you approach this limit without using the Squeeze Theorem?

**Tip:** The Squeeze Theorem is particularly useful when dealing with limits involving oscillatory functions, especially when the amplitude of oscillation becomes smaller as you approach the limit point.

Evaluate the limit \(\lim_{x \to 0} x \sin\left(\frac{9}{x}\right)\) using the Squeeze Theorem, given that \(-|x| \leq x \sin\left(\frac{9}{x}\right) \leq |x|\) for all \(x \neq 0\).

Learn how to evaluate the limit \(\lim_{x \to 0} x \sin\left(\frac{9}{x}\right)\) using the Squeeze Theorem. This solution walks you through the process of setting up bounds and applying the theorem to solve the limit problem. Ideal for college calculus students.

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