Computing $\lim_{x\to-\frac\pi2}\frac{e^{\tan x}}{\cos^2x}$ how can i compute: $\lim_{x\to-\frac\pi2}\frac{e^{\tan x}}{\cos^2x}$?
i tried l'hopital's rule  but it's like a loop.
also if it can be done without that rule i'd like to know how.
Thanks.

Hint. 
Let $x \to -\dfrac \pi2^-$, then
$$
\begin{align}
\tan x &=\frac{-\cos (x+\frac{\pi}{2})}{\sin (x+\frac{\pi}{2})} =\frac{-1}{x+\frac{\pi }{2}}+O\left(x+\frac{\pi }{2}\right)
\\\\\cos^2 x&=\sin^2 (x+\frac{\pi}{2})\sim\left(x+\frac{\pi}{2}\right)^2
\end{align}
$$ giving, as $x \to -\dfrac \pi2^-$,
$$
\frac{e^{\tan x}}{\cos^2x} \sim \frac{e^{\large -\frac{1}{x+\frac{\pi }{2}}}}{\left(x+\frac{\pi}{2}\right)^2} \to \infty\, \left(=\text{"}\frac{e^{+\infty}}{0^+}\text{"} \right).
$$
Let $x \to -\dfrac \pi2^+$, then
$$
\begin{align}
\tan x &=\frac{-\cos (x+\frac{\pi}{2})}{\sin (x+\frac{\pi}{2})} =\frac{-1}{x+\frac{\pi }{2}}+O\left(x+\frac{\pi }{2}\right)
\\\\\cos^2 x&=\sin^2 (x+\frac{\pi}{2})\sim\left(x+\frac{\pi}{2}\right)^2
\end{align}
$$ giving, as $x \to -\dfrac \pi2^+$,
$$
\frac{e^{\tan x}}{\cos^2x} \sim \frac{e^{\large -\frac{1}{x+\frac{\pi }{2}}}}{\left(x+\frac{\pi}{2}\right)^2} \to 0\, \left(=\text{"}\frac{e^{-\infty}}{0^+}\text{"} \right).
$$
