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$$\lim_{n \rightarrow \infty} \dfrac{(sin(\dfrac{1}{n}))^2}{n^2})$$

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.

Bee
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    You know $|\sin(x)| \leq 1$ so $|\sin^2(1/n)/n^2| \leq 1/n^2$. This is enough to prove the limit, but we can find a better bound by using $|\sin(x)|\leq |x|$ to find $|\sin^2(1/n)/n^2| \leq 1/n^4$. – Winther Jan 23 '15 at 06:53

7 Answers7

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I think the squeeze theorem will work nicely here.

user67953840
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Set $\dfrac1n=h$ to get $$\lim_{h\to0}h^2\sin^2h=0^2\cdot\sin^20=\cdots$$

lab bhattacharjee
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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?

user21820
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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.

curious_math
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$\begin{array}{lclrcl} youcansay\frac{1}{n} \to 0& & & & & \\ and f(x)=\frac{(sin(\frac{1}{n} ))^{2}}{n^{2}} =(sin(\frac{1}{n} ))^{2}& \times \frac{1}{n^{2}} & & & & \\ \Rightarrow \lim \limits_{\frac{1}{n} \to 0}f(x)=(sin(0))^{2}\times 0=0& & & & & \\ & & & & & ~ \end{array}$

SmartCoder
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$$\lim\limits_{n\to\infty} \frac{\sin^2\left(\frac{1}{n}\right)}{n^2}$$ Let $h=\frac1n$, then $$\lim\limits_{h\to 0} h^2\sin^2\left(h\right)=0\cdot 0=0$$

k170
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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.

Shocky2
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