Question

$lim_{x to 1} (2-x)^{tgfrac{pi}{2}x}$

Answer 1

Имеем неопределенность вида $1^{infty}$

 

$lim_{x->1} (2-x)^{tg frac{pi}{2}x}=\\|1-x=t; t->0, x=1-t|=\\ lim_{t->0} (1+t)^{tg(frac {pi}{2}-frac{pi}{2}t)}=\\ lim_{t->0} (1+t)^{ctg frac{pi}{2}t}=\\ lim_{t->0} (1+t)^{frac{1}{tg frac{pi}{2}t}}=\\ lim_{t->0} (1+t)^{frac{frac{pi}{2}t}{tg frac{pi}{2}t}*frac{2}{pi*t}}=\\ lim_{t->0} ((1+t)^frac{1}{t})^{frac{frac{pi}{2}t}{tg frac{pi}{2}t}*frac{2}{pi}}=\\ e^{1*frac{2}{pi}}=e^{frac{2}{pi}}$