Question

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

Answer 1

It depends...

Explanation:

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]}