# What are the removable and non-removable discontinuities, if any, of f(x)=x^2/absx?

Nov 2, 2015

$f$ has a removable discontinuity at $0$.

#### Explanation:

Recall that $\left\mid x \right\mid = \left\{\begin{matrix}x & \text{if" & x >= 0 \\ -x & "if} & x < 0\end{matrix}\right.$

So $f \left(x\right) = \left\{\begin{matrix}{x}^{2} / x & \text{if" & x > 0 \\ x^2/-x & "if} & x < 0\end{matrix}\right.$

Notice that $f$ is not defined for $x = 0$, so $f$ is not continuous at $0$.

We can simplify the definition of $f$, to get

$f \left(x\right) = \left\{\begin{matrix}x & \text{if" & x > 0 \\ -x & "if} & x < 0\end{matrix}\right.$

This is simply the absolute value function except that it is left undefined at $0$.

We know (I think) that the absolute value function is continuous, so $f$ is continuous everywhere except $0$.

Although $f \left(0\right)$ is not defined, we can see that ${\lim}_{x \rightarrow 0} f \left(x\right)$ exists. (It equals $0$), so the discontinuity at $0$ is removable.