$$\lim_{n \rightarrow \infty} \dfrac{(sin(\dfrac{1}{n}))^2}{n^2})$$
Steps I have taken:
Getting rid of the square through the limit of a product is the product of it's limit so I will square the limit at the end.
$\lim_{n \rightarrow \infty} \dfrac{sin(\dfrac{1}{n})}{n}$
$\lim_{n \rightarrow \infty} {sin(\dfrac{1}{n})}$ × $\lim_{n \rightarrow \infty} \dfrac{1}{n}$
I've now read a lot of posts here and on youtube about what the limit of sin($\dfrac{1}{n}$) and it seems that it equals 1 by comparing it to $\dfrac{1}{n}$ and using L'hopital's rule. My question here is how does one re-write this $\lim_{n \rightarrow \infty} {sin(\dfrac{1}{n})}$ to be $\lim_{n \rightarrow \infty} \dfrac{sin(\dfrac{1}{n})}{\dfrac{1}{n}}$ because I think I'm completely misunderstanding this step.
And to continue with the previous thought,
I now have the first part limit = 1 and the second limit = 0. So 1 × 0 = 0. And then back to the original step, $0^2$ = 0.
Best Answer
I think the squeeze theorem will work nicely here.