Trigonometric limit: $(1-\sqrt{\cos x})/x^2$ as $x\to 0$, without using L'Hopital I have to evaluate this limit without using L'Hopital. Could you help me $$\lim_{x \to 0} {1-\sqrt{\cos(x)}\over x^2}$$ I already rationalized it: $$\lim_{x \to 0} \left({1-\sqrt{\cos(x)}\over x^2}\right) \left({1+\sqrt{\cos(x)}\over 1+\sqrt{\cos(x)}}\right)$$ And I got: $$\lim_{x \to 0} \left({1-\cos(x)\over x^2(1+\sqrt{\cos(x)})}\right)$$ What should I do next?

Note that $$ \frac{1-\cos(x)}{x^2(1+\sqrt{\cos x})}=\frac{1-\cos^2 x}{x^2(1+\sqrt{\cos x})(1+\cos x)}=\left(\frac{\sin x}{x}\right)^2\frac{1}{(1+\sqrt{\cos x})(1+\cos x)} $$ then you can use $\lim_{x\to 0}\sin(x)/x=1$. Arguments for this latter fact can be seen here. Some of them do not employ L'Hopital's Rule.