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.

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 $\tan x^{\cos x} \to e^{0} = 1$
