Calculus: Early Transcendentals, Chapter 2, 2.2, Section 2.2, Problem 35

We can also put it in this way
lim_(x->2pi^-) xcsc(x)
=lim_(x->2pi^-) (x) * lim_(x->2pi^-) csc(x)
=(2pi) *(lim_(x->2pi^-) csc(x)) -----------------------(1)

As
lim_(x->2pi^-) csc(x) =lim_(x->2pi^-) 1/(sin(x)) = 1/(-0) = -oo
So, from (1) we get ,
(2pi) *(lim_(x->2pi^-) csc(x))
= (2pi) *(-oo)
= -oo is the infinite limit