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.
 A: 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 $x \to -\dfrac \pi2^+$,
$$
\frac{e^{\tan x}}{\cos^2x} \sim \frac{e^{\large -\frac{1}{x+\frac{\pi }{2}}}}{\left(x+\frac{\pi}{2}\right)^2} \to 0\, \left(=\text{"}\frac{e^{-\infty}}{0^+}\text{"} \right).
$$
A: It doesn't exist.  The denominator goes to zero and the numerator goes to either one or infinity because $lim_{x->-\pi/2}tan(x)$ doesn't exists.  It goes to infinity from the left and negative infinity from the right.  Therefore the top goes to either 0 or infinity and the limit goes to infinity.
A: $$\lim_{x\to-\frac\pi2}\frac{e^{\tan x}}{\cos^2x}$$
when x is close to $-\frac {\pi}{2}$, $$\sin(x) \lt -0.5$$
so $$\frac{e^{\tan x}}{\cos^2x}\lt \frac{e^{\frac {-0.5}{\cos x}}}{\cos^2x}$$ and we can use squeeze rule. 
let $t = \cos x$, we can transform the limit to,
$$\lim_{t\to 0}\frac{e^{\frac {-0.5}t}}{t^2}$$
let $y=e^{\frac {-0.5}t}$, $t=- \frac {0.5}{\ln y}$
$$\lim_{y\to 1}\frac{y}{(\frac {0.5}{\ln y})^2}=\lim_{y\to 1}\frac{y{(\ln y})^2}{0.25}=0$$
End
A: If you recognize some special limits in this expression, the rest is relatively easy.

*

*$\lim_{x\to \infty} xe^{-x}=0 $

*$\lim_{x\to 0} xe^{-x}=0$
Begin with the substitution: $$\frac{1}{\cos^2(x)}=\sec^2(x)=\tan^2(x)+1$$ Rewriting the limit expression: $$\lim_{x\to -\frac{\pi}{2}}(\tan^2(x)+1)\space e^{\tan(x)}=\lim_{x\to -\frac{\pi}{2}} \tan^2(x)\space e^{\tan(x)}\space + \space \lim_{x\to -\frac{\pi}{2}} e^{\tan(x)} $$ If we make the substitution $-\tan(x)=y$, we get: $$\lim_{y\to \infty} y^2 e^{-y}\space +\space \lim_{y\to \infty} e^{-y}=0+0=0 $$
