How can I argue that $\lim_{x \to 0} x^2 \cos\left(\frac{1}{x^2}\right) = 0$
I understand I have to use a squeeze theorem and that one piece goes to zero but I'm not sure how to tackle this problem to show on a test.
How can I argue that $\lim_{x \to 0} x^2 \cos\left(\frac{1}{x^2}\right) = 0$
I understand I have to use a squeeze theorem and that one piece goes to zero but I'm not sure how to tackle this problem to show on a test.
Use $-1 \le \cos(\frac{1}{x^2}) \le 1$ and multiply through by $x^2$. Since $x^2 \ge 0$, the inequalities remain valid.
We have that $-1 \leq \cos (1/x^2) \leq 1$ for any $x$. So $-x^2 \leq x^2\cos(1/x^2) \leq x^2$. Therefore $ \lim_{x \to 0} -x^2 \leq \lim_{x \to 0} x^2\cos(1/x^2) \leq \lim_{x \to 0} x^2. $ But we have that $ \lim_{x \to 0} x^2 = 0 $ and $ \lim_{x \to 0} -x^2 = 0. $
So $ 0 \leq \lim_{x \to 0} x^2\cos(1/x^2) \leq 0 $ and therefore by the squeeze theorem, $ \lim_{x \to 0} x^2\cos(1/x^2) = 0. $