# How do you find the x values at which f(x)=(x-1)/(x^2+x-2) is not continuous, which of the discontinuities are removable?

Oct 14, 2016

Discontinuity at x= -2

#### Explanation:

$\frac{x - 1}{{x}^{2} + 2 x - 2} = \frac{x - 1}{\left(x + 2\right) \left(x - 1\right)} = \frac{1}{x + 2}$
x=-2 is a vertical asymptote

Oct 14, 2016

Function $f$ is continuous at $a$ if and only if

${\lim}_{x \rightarrow a} f \left(x\right) = f \left(a\right)$.

There are three conditions in this equality:

$C 1 :$ $f \left(a\right)$ exists

$C 2 :$ ${\lim}_{x \rightarrow a} f \left(x\right)$ exists

$C 3 :$ The numbers in $C 1$ and $C 2$ are the same.

$f \left(x\right) = \frac{x - 1}{{x}^{2} + x - 2}$ is not defined for solutions to ${x}^{2} + x - 2 = 0$.

Therefore, $f$ is not continuous at $x = 1$ and at $x = - 2$

A discontinuity at $a$ is removable if the function is discontinuous at $a$, but the limit as $x$ approaches $a$ exists.

Checking the two discontinuities we found, we see that

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

Because this limit exists, the discontinuity at $1$ is removable.

${\lim}_{x \rightarrow - 2} f \left(x\right)$ does not exist. SO the discontinuity at $- 2$ is not removable. (In fact, it is an infinite discontinuity)

To remove the discontinuity at $1$, define a new function, $g$ by

$g \left(x\right) = \left\{\begin{matrix}f \left(x\right) & \text{ if " & x != 1" and " x!= -2 \\ 1/3 & " if } & x = 1\end{matrix}\right.$.

$g \left(x\right) = f \left(x\right)$ for all $x$ except $1$ (and $- 2$) and $g$ is continuous at $x = 1$.