# How do you find the x values at which f(x)=abs(x-3)/(x-3) is not continuous, which of the discontinuities are removable?

Feb 4, 2017

Refer to the Discussion given in the Explanation Section below.

#### Explanation:

Observe that $f$ is not defined at $x = 3$, and, hence is not

continuous at that point.

For AA x in (3,oo) ={ x in RR : x>3}; by the defn. of Absolute Value Function,

$| x - 3 | = \left(x - 3\right) \Rightarrow f \left(x\right) = | x - 3 \frac{|}{x - 3} = \frac{x - 3}{x - 3} = 1 , x > 3.$

$\text{Similarly, } \forall x \in \left(- \infty , 3\right) , f \left(x\right) = \frac{- \left(x - 3\right)}{x - 3} = - 1 , x < 3.$

This means that ${\lim}_{x \to 3 +} f \left(x\right) = 1 \ne - 1 = {\lim}_{x \to 3 -} f \left(x\right) .$

In other words, ${\lim}_{x \to 3} f \left(x\right) \text{ does not exist.}$

We conclude that $f \text{ has non-removable discontinuity at } x = 3.$

Feb 4, 2017

$f \left(x\right)$ is continuous in $\mathbb{R} - \left\{3\right\}$, and the dicontinuity is not removable.

#### Explanation:

The function is discontinuous for $x = 3$ since the denominator is null.

Then we analyze separately:

1. For $x \in \left(3 , + \infty\right)$, we have:
$\left(x - 3\right) > 0 \implies \left\mid x - 3 \right\mid = \left(x - 3\right) \implies f \left(x\right) = \frac{\left\mid x - 3 \right\mid}{x - 3} = \frac{x - 3}{x - 3} = 1$

2. For $x \in \left(- \infty , 3\right)$, we have:
$\left(x - 3\right) < 0 \implies \left\mid x - 3 \right\mid = - \left(x - 3\right) \implies f \left(x\right) = \frac{\left\mid x - 3 \right\mid}{x - 3} = - \frac{x - 3}{x - 3} = - 1$

Thus, $f \left(x\right)$ is continuous in $\mathbb{R} - \left\{3\right\}$

The discontinuity cannot be removed, as clearly:

${\lim}_{x \to {3}^{-}} f \left(x\right) = - 1$

${\lim}_{x \to {3}^{+}} f \left(x\right) = 1$

As the limit from the right and from the left are different, then the function does not have a limit for $x \to 3$

This is the graph of $y = f \left(x\right)$ which shows the discontinuity:
graph{(|x-3|)/(x-3) [-2, 5, -3, 3]}