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.