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

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


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



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

  • $\begingroup$ That last step is sort of an leap. You sort of need the step $$\dfrac{\tan(\pi \sin^2(x))}{x^2} = \pi\dfrac{\tan(\pi\sin^2(x))}{\pi\sin^2(x)}\dfrac{\sin^2(x)}{x^2}$$ $\endgroup$ – Thomas Andrews Mar 26 '13 at 5:49

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