Without using L'Hôpital's rule, find:
$$\lim_{x\to 0}\dfrac{\cos(\frac{\pi}{2}\cos x)}{\sin(\sin x)}$$
I know that the answer is $0$.
My attempt:
I tried by using the half-angle formula, $$\cos\left(\frac{A}{2}\right)=\pm\sqrt{\frac{1+\cos A}{2}},$$ but I've couldn't find it.
Best Answer
Rewrite the limit as a product like so:
$$L = \lim_{x\to 0} \frac{\sin x}{\sin( \sin x)} \cdot\frac{x}{\sin x} \cdot \frac{\cos(\frac{\pi}{2}\cos x)}{x}$$
If the limits exist individually then the limit of their product will be the product of their limits. The first two limits are $1$, giving us that
$$L = \lim_{x\to 0} \frac{\cos(\frac{\pi}{2}\cos x)}{x} = \lim_{x\to 0} \frac{\cos(\frac{\pi}{2}\cos x) - \cos(\frac{\pi}{2}\cos 0)}{x-0} \equiv \Bigr(\cos(\frac{\pi}{2}\cos x)\Bigr)'\Bigr|_{x=0} = 0$$
by the definition of the derivative.