# How do you find the limit of  (x - 3) / (abs(x - 3)) as x approaches 3?

Jul 21, 2016

${\lim}_{x \rightarrow 3} f \left(x\right)$ does not exist.

#### Explanation:

Recall that, $| y | = y , y \ge 0 , \mathmr{and} , | y | = - y , \mathmr{if} y < 0$.

Let us denote, by $f \left(x\right) = \frac{x - 3}{|} x - 3 | , x \ne 3$.

As $x \rightarrow 3 + , x > 3 \Rightarrow \left(x - 3\right) > 0 \Rightarrow | x - 3 | = x - 3$

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

$\therefore {\lim}_{x \rightarrow 3 +} f \left(x\right) = {\lim}_{x \rightarrow 3 +} 1 = 1. \ldots \ldots \ldots . \left(1\right)$

As $x \rightarrow 3 - , x < 3 \Rightarrow \left(x - 3\right) < 0 \Rightarrow | x - 3 | = - \left(x - 3\right)$

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

$\therefore {\lim}_{x \rightarrow 3 -} f \left(x\right) = {\lim}_{x \rightarrow 3 -} - 1 = - 1. \ldots \ldots \ldots \ldots . \left(2\right)$

From $\left(1\right) \mathmr{and} \left(2\right)$, we see that,

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

Evidently, ${\lim}_{x \rightarrow 3} f \left(x\right)$ does not exist.