# How do you find the limit of | x - 5 | as x approaches 5?

Oct 13, 2016

0

#### Explanation:

$\left\mid x - 5 \right\mid$ is a continuous funtion so ${\lim}_{x \to 5} \left\mid x - 5 \right\mid = \left\mid 5 - 5 \right\mid = 0$

Oct 13, 2016

If you are doing this to prove that the function is continuous, rewrite using the definition of absolute value.

#### Explanation:

$\left\mid u \right\mid = \left\{\begin{matrix}u & \text{ if " & u >= 0 \\ -u & " if } & u < 0\end{matrix}\right.$

With $u = x - 5$, the condition $u \ge 0$ becomes $x - 5 \ge 0$ which is equivalent to $x \ge 5$.
Similarly $u < 0$, becomes $x < 5$.

So,

$\left\mid x - 5 \right\mid = \left\{\begin{matrix}x - 5 & \text{ if " & x >= 5 \\ -(x-5) & " if } & x < 5\end{matrix}\right.$

${\lim}_{x \rightarrow {5}^{+}} \left\mid x - 5 \right\mid = {\lim}_{x \rightarrow {5}^{+}} \left(x - 5\right) = 0$
and
${\lim}_{x \rightarrow {5}^{-}} \left\mid x - 5 \right\mid = {\lim}_{x \rightarrow {5}^{-}} \left(- \left(x - 5\right)\right) = - \left(0\right) = 0$

Since both one-sided limits are $0$, the limit is $0$.