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

Feb 2, 2016

This function has a removable discontinuity at $x = 0$ and a non-removable discontinuity at $x = 2$.

#### Explanation:

When $x \ne 0$ and $x \ne 2$, we can write
$f \left(x\right) = \frac{x \left(x - 2\right)}{x \left(x - 2\right) \left(x - 2\right)} = \frac{\cancel{x} \left(\cancel{x - 2}\right)}{\cancel{x} \left(\cancel{x - 2}\right) \left(x - 2\right)} = \frac{1}{x - 2}$.

Therefore, the graph of $f$ is exactly the same as the graph of $g \left(x\right) = \frac{1}{x - 2}$ when $x \ne 0$ and $x \ne 2$. The graph of $g$ only has a discontinuity at $x = 2$, meaning the discontinuity of $f$ at $x = 2$ is non-removable (there's a vertical asymptote there).

On the other hand, $g \left(0\right) = - \frac{1}{2}$, meaning that ${\lim}_{x \to 0} f \left(x\right) = - \frac{1}{2}$, meaning the discontinuity of $f$ at $x = 0$ can be "removed" by re-defining $f$ in a piecewise way there to be $f \left(0\right) = - \frac{1}{2}$. In other words, the discontinuity of $f$ at $x = 0$ is removable.

The graph shown below is the graph of $g$, which is essentially the graph of the original (non-piecewise) $f$, except that the point $\left(x , y\right) = \left(0 , - \frac{1}{2}\right)$ would be "missing" from the graph of the original $f$.

graph{1/(x-2) [-10, 10, -5, 5]}