Limit of (sin(1/n)^2)/n^2 $$\lim_{n \rightarrow \infty} \dfrac{(sin(\dfrac{1}{n}))^2}{n^2})$$
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. 
 A: I think the squeeze theorem will work nicely here.
A: Set $\dfrac1n=h$ to get $$\lim_{h\to0}h^2\sin^2h=0^2\cdot\sin^20=\cdots$$
A: In general it isn't wise to attempt to use any rule or theorem without first understanding the behaviour. As $n \to \infty$, $\frac{1}{n} \to 0$, so what does $\sin(\frac{1}{n})$ go to? Likewise what does $n^2$ go to?
A: much simpler, by differentiating both numerator and denominator separately (there is a theorem for that) you get (-1/n^(3))*sin(1/n)cos(1/n). As you may see, this patern will continue so there is no point doing it. 
From the fraction 1/n we get 0 as n tend to infinity. So you have 0*0*1=0
PS: when n tends to infinity sin(1/n) tends to sin0 which as we know is 0.
A: $\begin{array}{lclrcl}
youcansay\frac{1}{n} \to 0& & & & & \\
and f(x)=\frac{(sin(\frac{1}{n} ))^{2}}{n^{2}} =(sin(\frac{1}{n} ))^{2}& \times \frac{1}{n^{2}} & & & & \\
\Rightarrow \lim \limits_{\frac{1}{n} \to 0}f(x)=(sin(0))^{2}\times 0=0& & & & & \\
& & & & & ~
\end{array}$
A: $$\lim\limits_{n\to\infty} \frac{\sin^2\left(\frac{1}{n}\right)}{n^2}$$
Let $h=\frac1n$, then
$$\lim\limits_{h\to 0} h^2\sin^2\left(h\right)=0\cdot 0=0$$
A: This question is really old, but I have an elegant solution: 
$sin^2(x) =  \frac{1-cos(2x)}{2} $ using this identity, we get 
$\lim_{n \to \infty}  \frac{1-cos(\frac{2}{n}) }{2n^2}$ we can separate
$\lim_{n \to \infty} \frac{1}{2^n} - \lim_{n \to \infty} \frac{cos(\frac{2}{n})}{2n^2}$
Here we got: 
$\lim_{n \to \infty} cos(\frac{2}{n}) = 0$ because $\lim_{n \to \infty}\frac{1}{n}=0$ and $cos(0)=1$
Then we have $\lim_{n \to \infty}\frac{1}{2n^2}$ which is equal to $0$.
Hope this helps, step by step.
