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To show that:

\[
\lim_{x \to 0} \left[ x^2 \left( \sin \frac{1}{x} + \cos \frac{1}{x} \right) \right] = 0
\]

we proceed as follows:

### Step 1: Analyze the behavior of \( \sin \frac{1}{x} \) and \( \cos \frac{1}{x} \) as \( x \to 0 \)
Both \( \sin \frac{1}{x} \) and \( \cos \frac{1}{x} \) are oscillatory functions as \( x \to 0 \). They do not tend to a specific limit, but they remain bounded because:
\[
-1 \leq \sin \frac{1}{x} \leq 1 \quad \text{and} \quad -1 \leq \cos \frac{1}{x} \leq 1.
\]
Thus, for any \( x \neq 0 \), both \( \sin \frac{1}{x} \) and \( \cos \frac{1}{x} \) are bounded within [-1, 1].

### Step 2: Analyze the term \( x^2 (\sin \frac{1}{x} + \cos \frac{1}{x}) \)
Since \( \sin \frac{1}{x} + \cos \frac{1}{x} \) is bounded, say:
\[
|\sin \frac{1}{x} + \cos \frac{1}{x}| \leq 2,
\]
we can multiply this by \( x^2 \), giving:
\[
|x^2 (\sin \frac{1}{x} + \cos \frac{1}{x})| \leq 2x^2.
\]

### Step 3: Take the limit as \( x \to 0 \)
Now, consider the limit:
\[
\lim_{x \to 0} 2x^2 = 0.
\]
Thus, by the Squeeze Theorem (since the expression \( x^2 (\sin \frac{1}{x} + \cos \frac{1}{x}) \) is squeezed between \( -2x^2 \) and \( 2x^2 \), both of which tend to 0 as \( x \to 0 \)), we conclude that:
\[
\lim_{x \to 0} \left[ x^2 \left( \sin \frac{1}{x} + \cos \frac{1}{x} \right) \right] = 0.
\]

### Conclusion:
The limit is indeed 0, as required.

Do you want more details or have any questions?

Here are 5 related questions for further exploration:
1. How does the Squeeze Theorem work in general cases?
2. What are other examples of limits involving oscillatory functions?
3. What is the behavior of \( \frac{\sin(1/x)}{x} \) as \( x \to 0 \)?
4. How can we apply L'Hôpital's Rule to more complex limits?
5. What role does boundedness play in solving limits involving oscillatory functions?

**Tip:** When dealing with oscillatory functions, always check if the expression is bounded, as this can often help simplify the limit using the Squeeze Theorem.

Show that the limit \( \lim_{x \to 0} \left[ x^2 \left( \sin \frac{1}{x} + \cos \frac{1}{x} \right) \right] = 0 \).

Learn how to solve the limit \( \lim_{x \to 0} \left[ x^2 \left( \sin \frac{1}{x} + \cos \frac{1}{x} \right) \right] = 0 \) using the Squeeze Theorem. This calculus problem involves oscillatory functions and requires understanding the behavior of sine and cosine functions as x approaches 0.

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