Calculus of a Single Variable, Chapter 3, 3.9, Section 3.9, Problem 14

If you evaluate the limit of function from the left, as x -> 0^- yields:
lim_(x->0^-) csc(2x) = lim_(x->0^-) 1/(sin(2x))
Replacing 2sinx*cosx for sin(2x) yields:
lim_(x->0^-) 1/(sin(2x)) = lim_(x->0^-) 1/(2sinx*cosx)
lim_(x->0^-) 1/(2sinx*cosx) = (1/2)lim_(x->0^-) 1/(sin x)*lim_(x->0^-) 1/(cos x)
Replacing 0^- for x yields:
(1/2)lim_(x->0^-) 1/(sin x)*lim_(x->0^-) 1/(cos x) = (1/2)*(1/(sin 0^-))*1/(cos 0^-)
(1/2)lim_(x->0^-) 1/(sin x)*lim_(x->0^-) 1/(cos x) = (1/2)*(-oo)*1 = -oo
If you evaluate the limit of function from the right, as x -> 0^+ yields:
lim_(x->0^+) csc(2x) = lim_(x->0^+) 1/(sin(2x))
lim_(x->0^+) 1/(sin(2x)) = (1/2)*(1/(sin 0^+))*1/(cos 0^+) = (1/2)*(+oo)*1 = +oo
Evaluating the limit of csc(2x) as x approaches to 0^- yields -oo , while evaluating the limit of csc(2x) as x approaches to 0^+ yields +oo, hence the two sided limit does not exist.