Limit question $\infty^{0}$ type $$\lim_{x\to\frac{\pi}{2}^-}  (\tan x)^{\cos x}$$
I just tried to write $e^{\ln(\tan x^{\cos x})}$ form but I couldn't solve the limit.
 A: We have: $\cos x\log(\tan x) = \dfrac{\log (\tan x)}{\sec x}$, and using L'hospital's rule:
$\dfrac{\dfrac{\sec^2 x}{\tan x}}{\sec x\tan x} = \dfrac{\sec x}{\sec^2 x - 1} = \dfrac{\cos x}{1 - \cos^2 x} \to 0$ as $x \to \pi/2^{-}$. Thus $\tan x^{\cos x} \to e^{0} = 1$
A: \begin{equation*}
\lim_{x \rightarrow 0}
(\cot x )^{\sin x}
\end{equation*}
is a more convenient expression for the same thing. Now 
\begin{equation*}
\lim_{x \rightarrow 0}
(\cos x )^{\sin x} = 1
\end{equation*} presents no problem. So take $u = \sin x$. We are left with
\begin{equation*}
\lim_{u \rightarrow 0}
(1/u)^{u}
\end{equation*}
This is easily handled by the tranformation you intended: $y = e^{-\ln u}$. Then
\begin{equation*}
\lim_{y \rightarrow \infty} 
e^{\ln{y}/y} \rightarrow 1
\end{equation*}
A: Change $x=\pi/2-t$: then the limit becomes
$$
\lim_{t\to0}(\cot t)^{\sin t}
$$
Take the logarithm:
$$
\lim_{t\to0}\sin t\log\cot t=\lim_{t\to0}(\sin t\log\cos t-\sin t\log\sin t)
$$
Can you go on by computing the limit of the two summands separately?
