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?

I have arrived at a formal argument using the definitions of limits.
Since $\lim\limits_{k\rightarrow0}\dfrac{\sin(k)}{k}=1$, for any $\epsilon>0$ there exists $\delta>0$ such that $\left|\dfrac{\sin(k)}{k}-1\right|<\epsilon$ for all $|k-0|<\delta$.
Substituting $\dfrac{1}{x^2}=k$:
$$
\left|\frac{\sin\left(\dfrac{1}{x^2}\right)}{\dfrac{1}{x^2}}-1\right|<\epsilon
$$
for all $\left|\dfrac{1}{x^2}\right|<\delta \Rightarrow \sqrt{\dfrac{1}{\delta}}<x$ (since $\delta>0$, $x>0$).
Therefore for all $\epsilon>0$, 
$$
\left|\frac{\sin\left(\dfrac{1}{x^2}\right)}{\dfrac{1}{x^2}}-1\right|<\epsilon
$$
for all $x>\sqrt{\dfrac{1}{\delta}}>0$ for some $\delta>0$. Therefore, by the definition of a limit at infinity:
$$
\lim_{x\rightarrow\infty}\frac{\sin\left(\dfrac{1}{x^2}\right)}{\dfrac{1}{x^2}}=1$$
The limit at $-\infty$ could similarly be arrived at using the other half of the absolute value.
