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

Mar 13, 2018

It depends...

#### Explanation:

A function $f \left(x\right)$ is continuous at a point $x = a$ in its domain if and only if:

${\lim}_{x \to a} f \left(x\right) \text{ }$ exists and is equal to $f \left(a\right)$.

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

By this definition, $f \left(x\right) = \csc 2 x$ is continuous everywhere.

The domain of $\csc 2 x = \frac{1}{\sin 2 x}$ is the set of values for which $\sin 2 x \ne 0$.

So $2 x \ne k \pi$ and hence $x \ne \frac{k \pi}{2}$ for any integer $k$.

Some authors would say that $f \left(x\right) = \csc 2 x$ is discontinuous at the points $x = \frac{k \pi}{2}$ on the grounds that:

${\lim}_{x \to {\left(k \pi\right)}^{+}} \csc 2 x = + \infty \ne - \infty = {\lim}_{x \to {\left(k \pi\right)}^{-}} \csc 2 x$

and:

${\lim}_{x \to {\left(\frac{\left(2 k + 1\right) \pi}{2}\right)}^{+}} \csc 2 x = - \infty \ne + \infty = {\lim}_{x \to {\left(\frac{\left(2 k + 1\right) \pi}{2}\right)}^{-}} \csc 2 x$

That is, the left and right limits disagree at the points $x = \frac{k \pi}{2}$. 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]}