Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top


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.


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

share|cite|improve this question
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
up vote 2 down vote accepted

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

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

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

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

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

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

share|cite|improve this answer
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


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

share|cite|improve this answer
Thank you so much! This really helped. :) – Siyanda Oct 22 '12 at 14:40

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.