To solve the limit

$\lim_{x \to 0} x^4 \cos\left(\frac{1}{x^2}\right) = 0$

using the Squeeze Theorem, we need to apply the theorem based on how the cosine function behaves. The Squeeze Theorem states that if $f(x) \leq g(x) \leq h(x)$ for all $x$ in some interval, and if $\lim_{x \to a} f(x) = \lim_{x \to a} h(x) = L$, then $\lim_{x \to a} g(x) = L$.

Step-by-Step Solution:

1. Bound the cosine function: We know that for all $x$, $\cos(\theta)$ is always between -1 and 1, regardless of the value of $\theta$. Thus, we have: $-1 \leq \cos\left(\frac{1}{x^2}\right) \leq 1$

2. Multiply by $x^4$: Multiply the entire inequality by $x^4$ (note that $x^4 \geq 0$ for all real $x$) to get: $-x^4 \leq x^4 \cos\left(\frac{1}{x^2}\right) \leq x^4$

3. Take the limit as $x \to 0$: Now, we take the limit of the outer bounds as $x \to 0$: $\lim_{x \to 0} (-x^4) = 0 \quad \text{and} \quad \lim_{x \to 0} x^4 = 0$

   Therefore, by the Squeeze Theorem: $\lim_{x \to 0} x^4 \cos\left(\frac{1}{x^2}\right) = 0$

Conclusion:

By the Squeeze Theorem, we have shown that the limit of $x^4 \cos\left(\frac{1}{x^2}\right)$ as $x \to 0$ is indeed 0.

Would you like further details or explanations?

Here are 5 related questions:

1. How is the Squeeze Theorem applied to more complex functions?
2. What happens to the limit if we replace $x^4$ with $x^2$ in the given expression?
3. Can you use the Squeeze Theorem when dealing with oscillating functions like sine and cosine?
4. How do we know that multiplying by $x^4$ preserves the bounds in this problem?
5. What other methods (besides the Squeeze Theorem) could be used to evaluate this limit?

Tip: When dealing with trigonometric functions in limits, bounding the function between known values is often a useful strategy for applying the Squeeze Theorem.