# Limit of $\frac{\sqrt{1-\cos x}}x$ using l'Hôpital

I'm trying to find $$\lim_{x\downarrow 0}\frac{\sqrt{1-\cos x}}{x}$$ using l'Hôpital's rule but I seem to be stuck in a loop. I have tried applying l'Hôpital several times but the derivatives always contain the radical and the whole thing ends up being $0\over0$

Is there some way to rewrite $1-\cos x$ or is there something else I'm missing?

• You might consider $\cos(x)= 1 - \dfrac{x^2}{2}+\cdots$ – Henry Nov 29 '15 at 10:39

HINT

I would say: $\quad\displaystyle \lim_{x\to0}\frac{\sqrt{1-\cos(x)}}{x}=\left(\lim_{x\to0}\frac{1-\cos(x)}{x^2}\right)^{1/2}=\cdots$

• @Henry: Thanks :) – georg Nov 29 '15 at 10:45
• You are welcome - it is a good step but originally had a typo – Henry Nov 29 '15 at 10:46

Hint:

$$1-\cos x=2 \sin^2 \frac{x}{2}.$$

Another hint : Multiply the top and the bottom by $\sqrt{1+\cos(x)}$.

You may observe that $$\sqrt{1-\cos x}=\sqrt{2\sin^2 (x/2)}=\sqrt{2}\:|\sin(x/2)|.$$

$1-\cos x=2\sin^2\frac{x}{2}$, so:

$$\lim_{x\to0}\frac{\sqrt{1-\cos x}}{x}=\lim_{x\to0}\frac{\sqrt{2}\big|\sin\frac{x}{2}\big|}{x}$$

$$=\lim_{x\to0}\frac{\frac{1}{\sqrt{2}}\big|\sin\frac{x}{2}\big|}{\frac{x}{2}}$$

$$=\frac{1}{\sqrt{2}}$$

• This supposes $x>0$ (which is the O.P.'s hypothesis anyway). – Bernard Nov 29 '15 at 10:47

$$\sqrt{1-\cos x}=\tan{\frac{x}{2}}\sqrt{1+\cos x}$$

$$\lim_{x\to0}\frac{\tan{\frac{x}{2}}\sqrt{1+\cos x}}{x}=\frac{\sqrt{2}}{2}=\frac{1}{\sqrt2}$$

• Another method is to use the Taylor series expansion – Claudeh5 Nov 29 '15 at 11:18