Calculus: Early Transcendentals, Chapter 4, Review, Section Review, Problem 12

You need to evaluate the limit, hence, you need to replace pi for x in expression under limit:
lim_(x->pi)(x-pi)csc x = (pi-pi)/(sin pi) = 0/0
Since the limit is indeterminate 0/0 , you may use l'Hospital's theorem:
lim_(x->pi)(x-pi)csc x =lim_(x->pi)(x-pi)/(sin x) = lim_(x->pi)((x-pi)')/((sin x)')
lim_(x->pi)((x-pi)')/((sin x)') = lim_(x->pi)(1)/(cos x)
Replacing by pi yields:
lim_(x->pi)(1)/(cos x) = 1/(cos pi) = 1/(-1) = -1
Hence, evaluating the limit, using l'Hospital's theorem, yields lim_(x->pi)(x-pi)csc x = -1.