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$\lim_{n\rightarrow\infty}\sqrt{n}\cos(\frac{\pi}{2}-\frac{1}{\sqrt{n}})$

I'm having a lot of trouble figuring it out. My first step is always to convert $\cos(\frac{\pi}{2}-\frac{1}{\sqrt{n}})$ to $\sin(-\frac{1}{\sqrt{n}})$ and then I get stuck here. Because I'm not quite sure where $\lim_{n\rightarrow\infty}\sqrt{n}(-\sin(\frac{1}{\sqrt{n}})$ leads....:/

Please help.

EDIT

My Thomas' Calculus text book (12th Edition) lists the identity as being $$cos(A-\frac{\pi}{2}) = sin(A)$$ so naturally (or perhaps, naively?) I went ahead and took my A to be $-\frac{1}{\sqrt{n}}$

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    If you know that $\lim\limits_{x\to0}\frac{\sin x}x=1$ - see [here](http://math.stackexchange.com/questions/75130/how-to-prove-that-lim-limits-x-to0-frac-sin-xx-1) - the rest should be easy. Also $\cos(\pi/2-x)=\sin x$; are you sure about your conversion?2012-10-22
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    Have you tried graphing this function on a calculator? That should at least help you guess what the answer should be.2012-10-22
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    The title is sweetly absurd.2012-10-22
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    @did *embarrassed*2012-10-22
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    @QiaochuYuan Yes, I'll use an app I have on google chrome and see where the graph leads me! Thanks :)2012-10-22

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$\cos(\frac\pi2-x)=\sin x$

$\cos(\frac\pi2-\frac 1{\sqrt n})=\sin \frac 1{\sqrt n}$

Put $h=\frac 1{\sqrt n},$ so, $h\to 0$ as $n\to ∞$

So, $\lim_{n\rightarrow\infty}\sqrt{n}\cos(\frac{\pi}{2}-\frac{1}{\sqrt{n}})$

$=\lim_{n\rightarrow\infty}\sqrt{n}\sin(\frac 1{\sqrt{n}})$

$=\lim_{ h\to 0}\frac{\sin h}{h}=1$

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    @Siyanda, welcome; hope I could clear the idea.2012-10-22
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Hint

$\lim\limits_{n\rightarrow\infty}\sqrt{n}\cos\left(\dfrac{\pi}{2}-\dfrac{1}{\sqrt{n}}\right)=\lim\limits_{n\rightarrow\infty}\sqrt{n}\sin\left(\dfrac{1}{\sqrt{n}}\right)=\lim\limits_{n\rightarrow\infty}\dfrac{\sin\left(\dfrac{1}{\sqrt{n}}\right)}{\dfrac{1}{\sqrt{n}}}.$ What did you know about $\lim\limits_{x\rightarrow 0}\dfrac{\sin{x}}{x}$?

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    Thank you so much! This really helped. :)2012-10-22