Single Variable Calculus, Chapter 7, 7.8, Section 7.8, Problem 62

Determine the $\displaystyle \lim_{x \to 1} (2-x)^{ \tan \left( \pi x/ 2\right) }$. Use L'Hospital's Rule where appropriate. Use some Elementary method if posible. If L'Hospitals Rule doesn't apply. Explain why.

If we let $\displaystyle y = (2-x)^{ \tan \left( \pi x/ 2\right) }$, then

$\displaystyle \ln y= \tan \left( \frac{\pi x}{2} \right) [ \ln (2-x)]$
So,
$\displaystyle \lim_{x \to 1} \ln y = \lim_{x \to 1} \tan \left( \frac{\pi x}{2} \right) [\ln (2-x)]$
$\displaystyle \lim_{x \to 1} \tan \left( \frac{\pi x}{2} \right) [\ln (2-x)] = \lim_{x \to 1} \frac{\ln(2-x)}{\cot \left( \frac{\pi x}{2} \right)}$

By applying L'Hospital's Rule...

$
\begin{equation}
\begin{aligned}
\lim_{x \to 1} \frac{\ln(2-x)}{\cot \left( \frac{\pi x}{2} \right)} &= \lim_{x \to 1} \frac{\frac{-1}{2-x}}{-\csc^2 \left( \frac{\pi x}{2} \right) \cdot \frac{\pi}{2}}\\
\\
&= \lim_{x \to 1} \frac{2}{\pi(2-x) \csc^2 \left( \frac{\pi x}{2} \right)}\\
\\
&= \frac{2}{\pi} \lim_{x \to 1} \frac{1}{(2-x)\left( \frac{1}{\sin^2\left( \frac{\pi x}{2} \right)} \right)}\\
\\
&= \frac{2 }{\pi} \lim_{x \to 1} \frac{\sin^2 \left( \frac{\pi x}{2} \right)}{(2 -x)}\\
\\
&= \frac{2}{\pi} \cdot \frac{\sin^2\left( \frac{\pi(1)}{2} \right)}{2-1}\\
\\
&= \frac{2}{\pi} \cdot \frac{\left( \sin \frac{\pi}{2} \right)^2}{1}\\
\\
&= \frac{2}{\pi} \cdot \frac{1}{1}\\
\\
&= \frac{2}{\pi}
\end{aligned}
\end{equation}
$

Thus, $\displaystyle \lim_{x \to 1} \ln y = \lim_{x \to 1} \tan \left(\frac{\pi x}{2} \right) [\ln(2-x)] = \frac{2}{\pi}$
Therefore, we have

$\displaystyle \lim_{x \to 1} (2 -x)^{\tan \left( \pi x/2 \right)} = \lim_{x \to 1} e^{\ln y} = e^{2/\pi}$