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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!


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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
up vote 3 down vote accepted

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 $\sin{\dfrac1n}$ – 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

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