TITLE: Find $\lim_{x\to 0} \frac{e^{\cos^2x}-e}{ \tan^2x}$
QUESTION [5 upvotes]: $$\lim_{x\to 0} \frac{e^{\cos^2x}-e}{ \tan^2x}=?$$

I'm at a complete loss here to be quite honest. I'm sure there is a way to simplify the $e$ part somehow. Factoring out the $e$ in the numerator to make $e(e^{-\sin^2x}-1)$ doesn't seem to lead anywhere.

REPLY [2 votes]: Put $\tan x=t$ so that $$\lim_{x\to 0} \frac{e^{\cos^2x}-e}{ \tan^2x}=\lim_{x\to 0} \frac{e^{\frac{1}{1+t^2}}-e}{t^2}$$
Applying Hôpital's rule $$\lim_{x\to 0} \frac{e^{\frac{1}{1+t^2}}-e}{t^2}=\lim_{x\to 0} \frac{-2te^{\frac{1}{1+t^2}}}{2t(1+t^2)^2}=\lim_{x\to 0} \frac{-e^{\frac{1}{1+t^2}}}{(1+t^2)^2}=-e$$