Single Variable Calculus, Chapter 2, Review Exercises, Section Review Exercises, Problem 18

Show that $\displaystyle \lim \limits_{x \to 0} x^2 \cos \left( \frac{1}{x^2} \right) = 0$

Proof:

$\qquad \lim \limits_{x \to 0} x^2 \cos \displaystyle \left( \frac{1}{x^2} \right) = \lim \limits_{x \to 0} x^2 \cdot \lim \limits_{x \to 0} \cos \displaystyle \left( \frac{1}{x^2} \right) $

$\qquad \lim \limits_{x \to 0} \cos \displaystyle \left( \frac{1}{x^2} \right)$ does not exist, the function is undefined because the denominator is equal to 0. However, since

$\qquad -1 \leq \cos \displaystyle \left( \frac{1}{x^2} \right) \leq 1$

We have,

$\qquad -x^2 \leq x^2 \cos \displaystyle \left( \frac{1}{x^2} \right) \leq x^2$

We know that

$\qquad \lim \limits_{x \to 0}-x^2 = -(0)^2 = 0$ and $\lim \limits_{x \to 0} x^2 = (0)^2 = 0$

Taking

$\qquad f(x) = -x^2, g(x) = x^2 \cos \displaystyle \left( \frac{1}{x^2} \right), h(x) =x^2$ in the Squeeze Theorem

We obtain,

$\qquad \displaystyle \lim \limits_{x \to 0} x^2 \cos \left( \frac{1}{x^2} \right) = 0$