The problem is to evaluate the following limit:

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

Solution:

1. Analyze the behavior of $\cos\left(\frac{1}{x}\right)$ as $x \to 0$:

   • The function $\cos\left(\frac{1}{x}\right)$ oscillates between $-1$ and $1$ as $x \to 0$.
   • This oscillation is unbounded in frequency as $x \to 0$, meaning it oscillates infinitely fast near zero.

2. Examine $x^2$ as $x \to 0$:

   • The term $x^2$ approaches zero as $x \to 0$.
   • So, the entire expression is the product of $x^2$ (which tends to zero) and $\cos\left(\frac{1}{x}\right)$ (which oscillates between $-1$ and $1$).

3. Applying the Squeeze Theorem:

   • We know that $-1 \leq \cos\left(\frac{1}{x}\right) \leq 1$ for all $x$.
   • Therefore, $-x^2 \leq x^2 \cos\left(\frac{1}{x}\right) \leq x^2$
   • As $x \to 0$, both $-x^2$ and $x^2$ approach $0$.
   • By the Squeeze Theorem, it follows that $\lim_{x \to 0} x^2 \cos\left(\frac{1}{x}\right) = 0.$

Final Answer:

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

Would you like further explanation on any part of this solution?

Here are some related questions:

1. What is the Squeeze Theorem and how does it apply to limits involving oscillating functions?
2. How does $x^n \cos\left(\frac{1}{x}\right)$ behave as $x \to 0$ for different powers $n$?
3. What happens if we change $x^2$ to $x$ in the limit $\lim_{x \to 0} x \cos\left(\frac{1}{x}\right)$?
4. Can the Squeeze Theorem be used to evaluate $\lim_{x \to 0} x^3 \cos\left(\frac{1}{x}\right)$?
5. How does the behavior of $\sin\left(\frac{1}{x}\right)$ compare to $\cos\left(\frac{1}{x}\right)$ in such limit problems?

Tip: When dealing with limits involving oscillating functions multiplied by terms that go to zero, consider using the Squeeze Theorem as a useful tool.