lim(2-x)tan(πx/2)
X->1
That's correct. Does this help you?
$\displaystyle
\mathop {\lim }\limits_{x \to 1} \left( {2 - x} \right)^{\tan \left( {\frac{{\pi x}}{2}} \right)} = \exp \left( {\mathop {\lim }\limits_{x \to 1} \left( {\tan \left( {\frac{{\pi x}}{2}} \right)\log \left( {2 - x} \right)} \right)} \right)
$
$\displaystyle
\mathop {\lim }\limits_{x \to 1} \left( {\tan \left( {\frac{{\pi x}}{2}} \right)\log \left( {2 - x} \right)} \right) = \mathop {\lim }\limits_{x \to 1} \frac{{\log \left( {2 - x} \right)}}{{\cot \left( {\frac{{\pi x}}{2}} \right)}}
$
Now you can apply l'Hôpital's rule.
We are interested in:
$\displaystyle
\lim_{x \to 1} ~(2-x)^ { \tan(\pi x/2)}
$
Put $\displaystyle y=x-1$, then:
$\displaystyle
~(2-x)^{ \tan(\pi x/2)}=(1-y)^{\cot(\pi y/2)}
$
but $\displaystyle \cot(\pi y/2) \approx 2/(\pi y)$ for small $\displaystyle y$
So as $\displaystyle y \to 0, \ \ (1-y)^{\cot(\pi y/2)} \to (1-y)^{\frac{2}{\pi y}}$
and with some jiggery pokery this last should come to $\displaystyle e^{2/pi}$ (it should rearrange into the classical limit form for an exponential)
RonL