Evaluate $ \lim_{x\rightarrow{\frac\pi2 }} (\sec(x) \tan(x))^{\cos(x)}$ without L'Hôpital's rule I have tried changing limit to $\lim_{x\rightarrow0}$ and use some trigonometry identity ($\sin^2(x)+\cos^2(x) = 1$ and $\sin (x+\pi/2) = \cos(x)$)  but doesn't work
I have no idea on how to do this now...
$$
\lim_{x\rightarrow{\frac\pi2 }} (\sec(x) \tan(x))^{\cos(x)} $$
 A: Hint: $$\lim_{t\to 0} \sin t \ln \frac{\cos t}{\sin^2t} = -2 \lim_{t\to 0} (\sin t \cdot \ln \sin t) = 0.$$
A: First observe that if $u=\cos x$: 
$$
\lim_{x\to \pi/2} (\sec x \tan x)^{\cos x}=\lim_{x\to \pi/2} \left(\frac{1}{\cos^2x}\right)^ {\cos x}=\lim_{u\to 0} \left(\frac{1}{u^2}\right)^ {u}
 $$
Now just assume that $L=\lim_{u\to 0} \left(\frac{1}{u}\right)^ {2u}$. You get:
$$
\ln L=\lim_{u\to 0} 2u\ln\left(\frac{1}{u}\right)=0 \implies L=1
$$

Remark: The last limit can be proved also without Hopital rule! To see this assume $y=\ln\left(\frac{1}{u}\right)$, then the limit turns out to be:
$$
\lim_{y\to\infty}2ye^{-y}
$$ 
To prove that the last limit is zero, we use the following inequality:
$$
0<2ye^{-y}\leq \frac{2y}{1+y+y^2/2}
$$
The limit of RHS goes to zero as $y\to\infty$.
A: $\lim_{x \to \pi/2}\left[\sec\left(x\right)\tan\left(x\right)\right]^{\cos\left(x\right)} = \lim_{x \to 0}x^{-2x} = \lim_{x \to 0}{\rm e}^{-2x\ln\left(x\right)}$.
Since
$\lim_{x \to 0}\left[-2x\ln\left(x\right)\right] = -2\lim_{x \to 0}{\ln\left(x\right) \over 1/x} = -2\lim_{x \to 0}{1/x \over -1/x^{2}} = 0$, $\displaystyle{\color{#ff0000}{\large\lim_{x \to \pi/2}\left[\sec\left(x\right)\tan\left(x\right)\right]^{\cos\left(x\right)} = 1}}$
$\newcommand{\abs}[1]{\left\vert #1\right\vert}$
${\bf\mbox{Without L'H$\hat{\rm o}$pital}}$:
$$
\abs{x\ln\left(x\right)}
\leq
\abs{x\left(x - 1\right)}
\to
0\quad\mbox{when}\quad x \to 0
$$
