# What are the removable and non-removable discontinuities, if any, of f(x)=(x^2-100) / (x+10) ?

##### 1 Answer
Mar 19, 2016

The only discontinuity is at $x = - 10$. It is removable.

#### Explanation:

$f \left(x\right) = \frac{{x}^{2} - 100}{x + 10}$ is a rational function.

Rational functions are continuous on their domains.

The domain of $f$ is $\left(- \infty , - 10\right) \cup \left(- 10 , \infty\right)$.

So, $f$ is continuous except at $x = - 10$.

The discontinuity is removable if and only if ${\lim}_{x \rightarrow - 10} f \left(x\right)$ exists. (Note that, "infinite limits" are limits that do not exist.)

${\lim}_{x \rightarrow - 10} \frac{{x}^{2} - 100}{x + 10}$ has initial form $\frac{0}{0}$ which is indeterminate. Reduce the ratio and try again.

${\lim}_{x \rightarrow - 10} \frac{{x}^{2} - 100}{x + 10} = {\lim}_{x \rightarrow - 10} \frac{\left(x + 10\right) \left(x - 10\right)}{x + 10}$

$= {\lim}_{x \rightarrow - 10} \left(x - 10\right) = - 20$

The limit exists, so the discontinuity is removable.