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?
 A: 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
$$\lim_{x\to\infty }x^2sin\frac{1}{x^2}=\lim_{x\to \infty }\frac{\sin\frac{1}{x^2}}{\frac{1}{x^2}}=\lim_{x\to\infty }\frac{\frac{1}{x^2}+\frac{1}{x^2}\varepsilon(x)}{\frac{1}{x^2}}=\lim_{x\to \infty }(1+\varepsilon(x))=1.$$
A: 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.
A: 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.
