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

Question

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?

Answer 1

HINT

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

Answer 2

Hint:

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

Answer 3

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

Answer 4

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

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

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

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

Answer 5

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

Answer 6

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

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