# What are the removable and non-removable discontinuities, if any, of f(x)=(x^2 - 3x + 2)/(x^2 - 1) ?

Dec 10, 2015

Removable at $x = 1$ and non-removable at $x = - 1$

#### Explanation:

$f \left(x\right) = \frac{{x}^{2} - 3 x + 2}{{x}^{2} - 1}$ is a rational function.

Therefore, $f$ is continuous on its domain.

The domain of $f$ is $\mathbb{R} - \left\{- 1 , 1\right\}$.

So $f$ has discontinuities at $- 1$ and at $1$. (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 $x = 1$
${\lim}_{x \rightarrow 1} f \left(x\right) = {\lim}_{x \rightarrow 1} \frac{{x}^{2} - 3 x + 2}{{x}^{2} - 1}$.

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

lim_(xrarr1)f(x) = lim_(xrarr1)((x-2)(x-1))/((x+1)(x-1)

$= {\lim}_{x \rightarrow 1} \frac{x - 2}{x + 1} = - \frac{1}{2}$

Because the limit exists, the discontinuity is removable.

At $x = - 1$
${\lim}_{x \rightarrow - 1} f \left(x\right) = {\lim}_{x \rightarrow 1} \frac{{x}^{2} - 3 x + 2}{{x}^{2} - 1}$.

This limit has form $\frac{6}{0}$, so the limit does not exist and the discontinuity cannot be removed.