How do you find the x values at which #f(x)=csc 2x# is not continuous, which of the discontinuities are removable?

The answer to this question depends on your definition of continuity.

A function #f(x)# is continuous at a point #x=a# in its domain if and only if:

#lim_(x->a) f(x)" "# exists and is equal to #f(a)#.

If it is continuous at every point in its domain then according to at least one definition of continuity, we would say that #f(x)# is continuous and has no discontinuities.

By this definition, #f(x) = csc 2x# is continuous everywhere.

The domain of #csc 2x = 1/(sin 2x)# is the set of values for which #sin 2x != 0#.

So #2x != kpi# and hence #x != (kpi)/2# for any integer #k#.

Some authors would say that #f(x) = csc 2x# is discontinuous at the points #x=(kpi)/2# on the grounds that:

#lim_(x->(kpi)^+) csc 2x = +oo != -oo = lim_(x->(kpi)^-) csc 2x#

and:

#lim_(x->(((2k+1)pi)/2)^+) csc 2x = -oo != +oo = lim_(x->(((2k+1)pi)/2)^-) csc 2x#

That is, the left and right limits disagree at the points #x=(kpi)/2#. Hence if we consider these points of discontinuity then they are not removable.

Note however that these points are not part of the domain.

graph{csc(2x) [-10, 10, -5, 5]}