# What are the removable and non-removable discontinuities, if any, of f(x)=(3x^2-7x-6 )/ (x-3 ) ?

Nov 1, 2015

Removable discontinuity at $x = 3$.
No non-removable discontinuity.

#### Explanation:

$f \left(x\right) = \frac{3 {x}^{2} - 7 x - 6}{x - 3}$

$= \left(3 x + 2\right) \frac{x - 3}{x - 3}$

${\lim}_{x \to 3} f \left(x\right) = 8$

We have the only discontinuity at $x = 3$ and it is removable.

#### Explanation:

When dealing with quotients of continuous functions, discontinuities arise at points that nullify the denominator.
In our specific case $f \left(x\right) = \frac{3 {x}^{2} - 7 x - 6}{x - 3}$, so the denominator is null if $x = 3$. Then the domain of this function is the union of two open intervals D=]-infty,3[ cup ]3,+infty[ and at $x = 3$ we have a discontinuity.

Intuitively, a discontinuity is removable if we can extend the function to a continuous one by assigning a value to the point where the discontinuity occurs. More formally, if ${x}_{0}$ is a discontinuity of f:]a,x_0[ cup ]x_0,b[ to RR, the discontinuity is said to be removable if exists finite ${\lim}_{x \to {x}_{0}} f \left(x\right)$.

In our specific case we want to check if ${\lim}_{x \to 3} \frac{3 {x}^{2} - 7 x - 6}{x - 3}$ exists and if it's finite. Since $3$ is a root of the numerator, we conclude that $3 {x}^{2} - 7 x - 6$ can be factored out by $x - 3$. We have that $3 {x}^{2} - 7 x - 6 = \left(3 x + 2\right) \left(x - 3\right)$, so
${\lim}_{x \to 3} \frac{3 {x}^{2} - 7 x - 6}{x - 3} = {\lim}_{x \to 3} \frac{\left(3 x + 2\right) \left(x - 3\right)}{x - 3} = {\lim}_{x \to 3} \left(3 x + 2\right) = 11$

Since the limit exists and is finite, $x = 3$ is a removable discontinuity.