Single Variable Calculus, Chapter 7, Review Exercises, Section Review Exercises, Problem 78

Evaluate $\displaystyle \lim_{x \to (\pi/2)^-} (\tan x)^{\cos x}$



$
\begin{equation}
\begin{aligned}

\text{If we let } y =& (\tan x)^{\cos x}, \text{ then}
\\
\\
\ln y =& \cos x \ln (\tan x)
\\
\\
\text{So, } &
\\
\\
\lim_{x \to (\pi/2)^-} \ln y =& \lim_{x \to (\pi/2)^-} \cos x \ln (\tan x)
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\\
=& \lim_{x \to (\pi/2)^-} \cos x \cdot \lim_{x \to (\pi/2)^-} \ln (\tan x)
\\
\\
=& \cos \frac{\pi}{2} \cdot \lim_{x \to (\pi/2)^-} \ln (\tan x)

\end{aligned}
\end{equation}
$


By applying L' Hospitals Rule


$
\begin{equation}
\begin{aligned}

=& 0 \cdot \lim_{x \to (\pi/2)^-} \frac{\sec^2 x}{\tan x}
\\
\\
=& 0 \cdot \lim_{x \to (\pi/2)^-} \frac{\displaystyle \frac{1}{\cos^2 x}}{\displaystyle \frac{\sin x}{\cos x}}
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=& 0 \cdot \lim_{x \to (\pi/2)^-} \frac{1}{\sin x \cos x}

\end{aligned}
\end{equation}
$


If we evaluate the limit, we will still get an indeterminate form, so we must apply the L' Hospitals Rule once more..


$
\begin{equation}
\begin{aligned}

=& 0 \cdot \lim_{x \to (\pi/2)^-} 0
\\
\\
=& 0

\end{aligned}
\end{equation}
$


Thus,


$
\begin{equation}
\begin{aligned}

\lim_{x \to (\pi/2)^-} \ln y =& \lim_{x \to (\pi/2)^-} \cos x \ln = 0

\end{aligned}
\end{equation}
$


Therefore, we have..


$
\begin{equation}
\begin{aligned}

\lim_{x \to (\pi/2)^-} (\tan x)^{\cos x} =& \lim_{x \to (\pi/2)^-} e^{\ln y}
\\
\\
=& e^0
\\
\\
=& 1

\end{aligned}
\end{equation}
$