How to show $\lim _{x\to 0}\:\frac{\sin\left(\frac{1}{x^2}\right)}{x^2}$ does not exist? 
$$\lim _{x\to 0}\:\frac{\sin\left(\frac{1}{x^2}\right)}{x^2}$$

I my intuition is telling me this limit does not exist as $\sin$ will be oscillating but will stay bounded and then will blow up as $x \to 0$.
I can't seem to show it though. I tried to create two sequences that go to different limits but the $x^2$ in the numerator made this tricky so I'm not sure if that is intended?
Any help?
 A: Let's write $t=\frac{1}{x^2}$, such that when $x\rightarrow 0$, then $t \rightarrow \infty$. Let's notice that $\lim_{x\rightarrow 0} \frac{\sin(\frac{1}{x^2})}{x^2}=\lim_{t\rightarrow \infty}t\sin(t)$, which doesn't exist and therefore the original limit doesn't exist.
A: One may consider, $\dfrac1{x_n^2}=\pi n$, with $n=1,2,3,\ldots$, giving

$$
\lim _{x_n\to 0}\:\frac{\sin\left(\frac{1}{x_n^2}\right)}{x_n^2}=\lim _{n\to +\infty}\pi n \times \sin(\pi n)=0
$$ 

and one may consider, $\dfrac1{y_n^2}=(4n+1)\dfrac{\pi}2 $, with $n=0,1,2,\ldots$, giving

$$
\lim _{y_n\to 0}\:\frac{\sin\left(\frac{1}{y_n^2}\right)}{y_n^2}=\lim _{n\to +\infty} (4n+1)\dfrac{\pi}2 \times \sin\left((4n+1)\dfrac{\pi}2\right)=\lim _{n\to +\infty}(4n+1)\dfrac{\pi}2 \times 1=\infty.
$$

Thus the initial limit does not exist.
A: Make the substitution $n = \frac{1}{X^2}$. Then it is clear that $n \to \infty \iff x \to 0 $. HEnce, we have 
$$ \lim_{n \to \infty} n \sin n $$
Now, consider subsequences multiples of $pi$ and $pi/2$ and see that they converge to different limits...
