Evaluate: $\lim_{x \to 0}\left(\frac{\tan\left(\pi\cos^2x\right)}{x^2}\right)$

Question

How to evaluate the following limit without using L'Hospital's rule?

$$\lim_{x \to 0}\left(\frac{\tan\left(\pi\cos^2x\right)}{x^2}\right)$$

Answer 1

Depends on what you're willling to take

Heuristically $\cos(x)\sim 1-\frac{1}{2}x^2$ and so $\cos(x)^2\sim 1-x^2$. Thus, if $L$ denotes your limit:

$$L=\lim_{x\to 0}\frac{\tan(\pi-\pi x^2)}{x^2}=-\lim_{x\to 0}\frac{\tan(\pi x^2)}{x^2}=-\pi$$

where the last equality follows from substitution $\pi x^2\mapsto t$ and the (common) fact that

$$\lim_{t\to0}\frac{\tan(t)}{t}=1$$

Answer 2

$$\tan(\pi \cos^2(x)) = \tan(\pi - \pi \sin^2(x)) = - \tan(\pi \sin^2(x))$$ Hence, $$\lim_{x \to 0} \dfrac{\tan(\pi \cos^2(x))}{x^2}=- \lim_{x \to 0} \dfrac{\tan(\pi \sin^2(x))}{x^2} = - \pi$$