Evaluate the limit.
![\[ \lim_{x \to 1} (2-x)^{\tan \left( \frac{\pi x}{2} \right) }. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-f89dc1a97a2ab0b1bbe23238d017063a_l3.png "Rendered by QuickLaTeX.com")
First, we use the continuity and definition of the exponential,
![\begin{align*} \lim_{x \to 1} (2-x)^{\tan \left( \frac{\pi x}{2} \right)} &= \lim_{x \to 1} e^{ \tan \left( \frac{\pi x}{2} \right) \cdot \log (2-x) } \\[9pt] &= \exp \left( \lim_{x \to 1} \tan \left( \frac{\pi x}{2} \right) \log (2-x) \right). \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-93bcd2371006d94373e75518ad03b87c_l3.png "Rendered by QuickLaTeX.com")
Next, we use that 
to rewrite the term in the limit in the indeterminate form 
and apply L’Hopital’s rule,
![\begin{align*} \exp \left( \lim_{x \to 1} \tan \left( \frac{\pi x}{2} \right) \log (2-x) \right) &= \exp \left( \lim_{x \to 1} \frac{\log (2-x)}{\cot \frac{\pi x}{2} } \right) \\[9pt] &= \exp \left( \lim_{x \to 1} \frac{-\frac{1}{2-x}}{-\frac{\pi}{2} \frac{1}{\sin^2 \frac{\pi x}{2}}} \right) \\[9pt] &= \exp \left( \lim_{x \to 1} \frac{2 \sin^2 \left( \frac{\pi x}{2} \right)}{\pi(2-x)} \right) \\[9pt] &= e^{\frac{2}{\pi}}. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-6a4295a4f35c59789963b145131b84e9_l3.png "Rendered by QuickLaTeX.com")
Evaluate the limit.
First, we use the continuity and definition of the exponential,
Next, we use that to rewrite the term in the limit in the indeterminate form and apply L’Hopital’s rule,