The value of $\underset{x\rightarrow 1}{lim}(2-x)^{tan\left(\dfrac{\pi x}{2}\right)}$ is

Let $L=\lim _{x \rightarrow 1}(2-x)^{\tan \left(\frac{\pi x}{2}\right)}$. Taking the natural logarithm of both sides:

$\ln L=\lim _{x \rightarrow 1} \tan \left(\frac{\pi x}{2}\right) \cdot \ln (2-x)$

$\ln L=\lim _{x \rightarrow 1} \tan \left(\frac{\pi x}{2}\right) \cdot \ln (2-x)$

$\ln L=\lim _{x \rightarrow 1} \frac{\ln (2-x)}{\cot \left(\frac{\pi x}{2}\right)}$

As $x \rightarrow 1$, both the numerator and denominator approach 0, so we can apply L'Hôpital's Rule.

Differentiate the numerator and the denominator with respect to $x$ :

$\begin{array}{c} \frac{d}{d x} \ln (2-x)=-\frac{1}{2-x} \\ \frac{d}{d x} \cot \left(\frac{\pi x}{2}\right)=-\frac{\pi}{2} \csc ^{2}\left(\frac{\pi x}{2}\right) \end{array}$

Now, apply L'Hôpital's Rule:

$\ln L=\lim _{x \rightarrow 1} \frac{-\frac{1}{2-x}}{-\frac{\pi}{2} \csc ^{2}\left(\frac{\pi x}{2}\right)}=\lim _{x \rightarrow 1} \frac{2}{\pi(2-x) \cosec ^{2}\left(\frac{\pi x}{2}\right)}$

so the overall expression approaches:

$\ln L=\frac{2}{\pi}$

$L=e^{\frac{2}{\pi}}$