Calculus: Early Transcendentals, Chapter 4, 4.4, Section 4.4, Problem 64

You need to evaluate the limit, hence, you need to replace 1 for x:
lim_(x->1) (2 - x)^(tan((pi*x)/2)) = (2-1)^oo = 1^oo
You need to use the logarithm special technique but first you need to define the followings:
f(x) = (2 - x)^(tan((pi*x)/2))
Taking logarithms both sides yields:
ln f(x) = ln (2 - x)^(tan((pi*x)/2))
ln f(x) = ((tan((pi*x)/2))) * ln (2 - x)
Taking the limit:
lim_(x->1) ln f(x) = lim_(x->1)((tan((pi*x)/2))) * ln (2 - x) = oo*0
lim_(x->1)((tan((pi*x)/2))) * ln (2 - x) = lim_(x->1) (ln (2 - x))/(1/((tan((pi*x)/2))) = 0/0
You may use that 1/((tan((pi*x)/2))) = cot((pi*x)/2)
lim_(x->1) (ln (2 - x))/(cot ((pi*x)/2))
You may use l'Hospital's rule:
lim_(x->1) (ln (2 - x))/(cot ((pi*x)/2)) = lim_(x->1) ((ln (2 - x))')/((cot ((pi*x)/2))')
lim_(x->1) ((ln (2 - x))')/((cot ((pi*x)/2))') = lim_(x->1) (-1/(2-x))/(-(pi/2)csc^2((pi*x)/2)) = (-1/1)/(-pi/2) = 2/pi
Hence, evaluating the limit, yields lim_(x->1) (2 - x)^(tan((pi*x)/2)) = e^(2/pi).