Limit of $\cos(1/x^2)$ as $x$ tends to $0$

Question

$\displaystyle\lim_{x\to0}\cos\frac{1}{x^2}$

I understand that through the squeeze theorem we know the upper bound is $1$ and lower bound is $-1$

I understand that it would be undefined at $0$ and that there would be an infinite number of oscillations as $x$ tends to zero, but then what after this?

I cannot seem to crack this one.

Answer 1

HINT

Let consider

$$x_n=\frac1{\sqrt{\pi n}}$$

and take the limit $\cos (1/x_n^2)$ for $n$ odd and for $n$ even.