Имеем неопределенность вида 
1} (2-x)^{tg \frac{\pi}{2}x}=\\\\|1-x=t; t->0, x=1-t|=\\\\ lim_{t->0} (1+t)^{tg(\frac {\pi}{2}-\frac{\pi}{2}t)}=\\\\ lim_{t->0} (1+t)^{ctg \frac{\pi}{2}t}=\\\\ lim_{t->0} (1+t)^{\frac{1}{tg \frac{\pi}{2}t}}=\\\\ lim_{t->0} (1+t)^{\frac{\frac{\pi}{2}t}{tg \frac{\pi}{2}t}*\frac{2}{\pi*t}}=\\\\ lim_{t->0} ((1+t)^\frac{1}{t})^{\frac{\frac{\pi}{2}t}{tg \frac{\pi}{2}t}*\frac{2}{\pi}}=\\\\ e^{1*\frac{2}{\pi}}=e^{\frac{2}{\pi}}" alt="lim_{x->1} (2-x)^{tg \frac{\pi}{2}x}=\\\\|1-x=t; t->0, x=1-t|=\\\\ lim_{t->0} (1+t)^{tg(\frac {\pi}{2}-\frac{\pi}{2}t)}=\\\\ lim_{t->0} (1+t)^{ctg \frac{\pi}{2}t}=\\\\ lim_{t->0} (1+t)^{\frac{1}{tg \frac{\pi}{2}t}}=\\\\ lim_{t->0} (1+t)^{\frac{\frac{\pi}{2}t}{tg \frac{\pi}{2}t}*\frac{2}{\pi*t}}=\\\\ lim_{t->0} ((1+t)^\frac{1}{t})^{\frac{\frac{\pi}{2}t}{tg \frac{\pi}{2}t}*\frac{2}{\pi}}=\\\\ e^{1*\frac{2}{\pi}}=e^{\frac{2}{\pi}}" align="absmiddle" class="latex-formula">