Limit question $\infty^{0}$ type

Question

$$\lim_{x\to\frac{\pi}{2}^-} (\tan x)^{\cos x}$$

I just tried to write $e^{\ln(\tan x^{\cos x})}$ form but I couldn't solve the limit.

Answer 1

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

Answer 2

\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*}

Answer 3

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?