# What are the removable and non-removable discontinuities, if any, of f(x)=(x^4+4x^2+2x)/(x^2-2x)?

Dec 11, 2015

There is a removable discontinuity at $0$ and a non-removable discontinuity at $2$.

#### Explanation:

$f \left(x\right) = \frac{{x}^{4} + 4 {x}^{2} + 2 x}{{x}^{2} - 2 x}$ is a rational function.

Therefore, $f$ is continuous on its domain.

The domain of $f$ is $\mathbb{R} - \left\{0 , 2\right\}$.

So $f$ has discontinuities at $0$ and at $2$. (It is continuous everywhere else.)

In order to determine whether a discontinuity is removable, we need to look at the limit of $f$ as $x$ approaches the discontinuity.

At $x = 0$
${\lim}_{x \rightarrow 0} f \left(x\right) = {\lim}_{x \rightarrow 0} \frac{{x}^{4} + 4 {x}^{2} + 2 x}{{x}^{2} - 2 x}$.

This limit has indeterminate form $\frac{0}{0}$, so we factor and simplify.

${\lim}_{x \rightarrow 0} f \left(x\right) = {\lim}_{x \rightarrow 0} \frac{x \left({x}^{3} + 4 x + 2\right)}{x \left(x - 2\right)}$

$= {\lim}_{x \rightarrow 0} \frac{{x}^{3} + 4 x + 2}{x - 2} = - 1$

Because the limit exists, the discontinuity is removable.

At $x = 2$
${\lim}_{x \rightarrow 2} f \left(x\right) = {\lim}_{x \rightarrow 2} \frac{{x}^{3} + 4 {x}^{2} + 2 x}{{x}^{2} - 2 x}$.

This limit has form $\frac{28}{0}$, so the limit does not exist and the discontinuity cannot be removed.