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?

Your way is perfect ! 
An other way : 
Since $$\sin\frac{1}{x^2}=\frac{1}{x^2}+\frac{1}{x^2}\varepsilon(x)$$
where $\varepsilon(x)\to 0$ when $x\to\infty $, you get
$$\lim_{x\to\infty }x^2sin\frac{1}{x^2}=\lim_{x\to \infty }\frac{\sin\frac{1}{x^2}}{\frac{1}{x^2}}=\lim_{x\to\infty }\frac{\frac{1}{x^2}+\frac{1}{x^2}\varepsilon(x)}{\frac{1}{x^2}}=\lim_{x\to \infty }(1+\varepsilon(x))=1.$$
