# Squeeze theorem proof of $\lim_{n \rightarrow \infty} \sin{\frac{1}{n}}$

I've been having a bit of trouble with a tutorial question from my 3rd year Foundations of Calculus course.

It asks to find the limit of the sequence $\sin{\frac{1}{n}}$ as n approaches infinity, using the pinching theorem. I know the limit must be 0, but I'm not quite sure how to get there using the theorem.

I've got to $\frac{-1}{n} \leq \frac{1}{n} \sin{\frac{1}{n}} \leq \frac{1}{n}$ but can't see a clear way to go from there to just $\sin{\frac{1}{n}}$.

This may be because I've been staring at it too long and am missing something obvious, so it would be lovely if someone could point me in the right direction!

Thanks!

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Sorry, the question is actually $\lim_{n \rightarrow \infty} \sin{\frac{1}{n}}$, not $\frac{1}{n}\sin{\frac{1}{n}}$ (sorry, I realise that was a little confusing ...) – vim Apr 6 '13 at 11:12
There is a very useful inequality: $|\sin x| \leq |x|$ for all $x\in\Bbb R$. – Siméon Apr 6 '13 at 11:12

Hint: $$\sin x\le x \ \ \forall x>0$$ and $$\sin x\ge x\ \ \forall x<0$$

Or in other words: $$|\sin x|\le |x|$$

You can use $$\lim_{h\to 0^+}\ \ \Big ( -h\le \sin h \le h \Big)$$

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Ah gosh, I knew it was something horribly simple! My excuse is I'm a physics student and haven't done pure math for at least 2 years :P. – vim Apr 6 '13 at 11:13
It's ok , though simple sometimes We get stuck if mind is exhausted or we are not in touch. – ABC Apr 6 '13 at 11:15
you should add some absolute values else both inequalites are wrong – Dominic Michaelis Apr 6 '13 at 11:17
@vim See this for information about $\sin{\dfrac1n}$ :wolframalpha.com/input/?i=+sin%281%2Fn%29 – ABC Apr 6 '13 at 11:18
@Ju'x I already told him that his inequalites are wrong when he doesn't use absolute values, but seems like he don't care – Dominic Michaelis Apr 6 '13 at 11:24

If $n>\frac{2}{\pi}$, $0<\frac{1}{n}<\frac{\pi}{2}$ hence $0<\sin(\frac{1}{n})<\frac{1}{n}$.

As $n\to \infty$, $\frac{1}{n}\to 0$. Thus, by the squeeze theorem $\sin\dfrac1n\to0$

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How you got from first range of 1/n to sin(1/n)<1/n ? – ABC Apr 6 '13 at 11:15
$\sin(x)<x$ for $x>0$. You can convince yourself by looking at the graph for `large' x and looking at the taylor expansion near $0$ for instance. – Abel Apr 6 '13 at 11:17
I'm assuming by taking the sine of both sides? Thanks, this is a cool way to do it that I wouldn't have thought of straight away. – vim Apr 6 '13 at 11:20
That's what i have used in my solution. But what you have done in 1st line after $0<1/n<\pi/2$ to go to $0<sin(1/n)<1/n$ How? @vim By taking sine on both side we get $0<sin(1/n)<1$ – ABC Apr 6 '13 at 11:21
Ah yes of course, sorry. Clearly I shouldn't be doing this past midnight ... – vim Apr 6 '13 at 11:23

Well it's not exactly the squeeze theorem (at least not in the first step) but we notice that when we set $x=\frac{1}{n}$ that $$\lim_{n\to \infty}f\left(\frac{1}{n}\right)=\lim_{x\to 0^+ } f(x)$$ So we see that $$\lim_{n\to \infty} \sin\left(\frac{1}{n}\right) = \lim_{x\to 0^+} \sin(x)$$ And here we can use that for $x>0$. $$-x\leq \sin(x)\leq x$$

The substitution is surely not necessary, but I think using the substitution makes it clearer.

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Of course, the equality $$\lim_{n\to \infty}\frac{1}{n} f\left(\frac{1}{n}\right)=\lim_{x\to 0^+ } f(x)$$ is false: take $f(x)=\frac{1}{x}$ as counter-example. – Taladris Apr 6 '13 at 11:40
Dominic the 1/n is obviously wrong. – ABC Apr 6 '13 at 11:43
oh lol yeah the $\frac{1}{n}$ is bullshit sry for that wanted to write something else first i gonna fix that – Dominic Michaelis Apr 6 '13 at 11:49
@exploringnet fixed it – Dominic Michaelis Apr 6 '13 at 11:55
@user70123 fixed and thanks again i should proofread my proofs ;) – Dominic Michaelis Apr 6 '13 at 11:55

We have $\sin x=_0O(x)$ so $$\left|\sin\left(\frac{1}{n}\right)\right|\leq\frac{C}{n}$$

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Hey OP know this but he want to use Sandwich rule. – ABC Apr 6 '13 at 11:53
@exploringnet You're right I misread the question. – user63181 Apr 6 '13 at 12:05
Yes. you did read wrong. – ABC Apr 6 '13 at 12:06