Swapping limit at infinity with limit at 0 I am trying to calculate:
$$\lim_{x\rightarrow\infty}x^2\sin\left(\frac{1}{x^2}\right)
$$
I am pretty sure that this is equivalent to calculating:
$$\lim_{k\rightarrow0}\frac{\sin(k)}{k}=1
$$
Since $k=\dfrac{1}{x^2}$ and $\lim\limits_{x\rightarrow\infty}k=0$.
Is there any way I can make this formal?

Yes,state $$k=\frac{1}{x^{2}}$$
Then we have, $$ x^2=\frac{1}{k}$$
As $x\rightarrow\infty$, $k\rightarrow0$ so, 
$$lim_{k\rightarrow0}\frac{\sin(k)}{k}= 1$$
Since it is perfect!!
If you want to make it more formal simply use l'hospitals rule.
