TITLE: Evaluate: $\lim_{x \to 0}\left(\frac{\tan\left(\pi\cos^2x\right)}{x^2}\right)$
QUESTION [5 upvotes]: How to evaluate the following limit without using L'Hospital's rule?
$$\lim_{x \to 0}\left(\frac{\tan\left(\pi\cos^2x\right)}{x^2}\right)$$

REPLY [3 votes]: $$\frac{\tan(\pi\cos^2x)}{x^2}=\frac{\tan(\pi-\pi\sin^2x)}{x^2}=-\frac{\pi}{\cos(\pi\sin^2x)}\frac{\sin(\pi\sin^2x)}{\pi\sin^2x}\frac{\sin^2x}{x^2}\to-\frac\pi1\cdot1\cdot1.$$