# What is the limit of this function as $n$ tends to infinity?

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

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....:/

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 $-\frac{1}{\sqrt{n}}$

-
If you know that $\lim\limits_{x\to0}\frac{\sin x}x=1$ - see here - the rest should be easy. Also $\cos(\pi/2-x)=\sin x$; are you sure about your conversion? –  Martin Sleziak Oct 22 '12 at 14:18
Have you tried graphing this function on a calculator? That should at least help you guess what the answer should be. –  Qiaochu Yuan Oct 22 '12 at 14:19
The title is sweetly absurd. –  Did Oct 22 '12 at 14:49
@did embarrassed –  Siyanda Oct 22 '12 at 15:08
@QiaochuYuan Yes, I'll use an app I have on google chrome and see where the graph leads me! Thanks :) –  Siyanda Oct 22 '12 at 15:13

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

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

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

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

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

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

-
Thank you! Great method! –  Siyanda Oct 22 '12 at 14:41
@Siyanda, welcome; hope I could clear the idea. –  lab bhattacharjee Oct 22 '12 at 14:43

# 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 $\lim\limits_{x\rightarrow 0}\dfrac{\sin{x}}{x}$?

-
Thank you so much! This really helped. :) –  Siyanda Oct 22 '12 at 14:40