Limit of $\frac{\sqrt{1-\cos x}}x$ using l'Hôpital
Solution 1:
HINT
I would say: $\quad\displaystyle \lim_{x\to0}\frac{\sqrt{1-\cos(x)}}{x}=\left(\lim_{x\to0}\frac{1-\cos(x)}{x^2}\right)^{1/2}=\cdots$
Solution 2:
Hint:
$$1-\cos x=2 \sin^2 \frac{x}{2}. $$