# How do you find the limit of |x+2| / (x+2) as x approaches -2?

Apr 16, 2016

The limit does not exist.

#### Explanation:

The absolute value function $\left\mid x + 2 \right\mid$ can be defined as the piecewise function

abs(x+2)={(x+2,;,x>=-2),(-(x+2),;,x<-2):}

We should determine if the limit from the left approaches the limit from the right.

Limit from the left:

When the function is directly to the left of $x = - 2$, we are on the $- \left(x + 2\right)$ portion of the piecewise function since $x < - 2$.

Thus, the function when $x < - 2$ becomes

$\frac{\left\mid x + 2 \right\mid}{x + 2} = \frac{- \left(x + 2\right)}{x + 2} = - 1$

Hence the limit from the left is

${\lim}_{x \rightarrow - {2}^{-}} \frac{\left\mid x + 2 \right\mid}{x + 2} = {\lim}_{x \rightarrow - {2}^{-}} - 1 = - 1$

Limit from the right:

From the right, $x > - 2$, so we just use $x + 2$ in place of $\left\mid x + 2 \right\mid$.

The function becomes

$\frac{\left\mid x + 2 \right\mid}{x + 2} = \frac{x + 2}{x + 2} = 1$

So the limit from the right is

${\lim}_{x \rightarrow - {2}^{+}} \frac{\left\mid x + 2 \right\mid}{x + 2} = {\lim}_{x \rightarrow - {2}^{+}} 1 = 1$

Relating the limits:

Since ${\lim}_{x \rightarrow - {2}^{+}} \frac{\left\mid x + 2 \right\mid}{x + 2} \ne {\lim}_{x \rightarrow - {2}^{-}} \frac{\left\mid x + 2 \right\mid}{x + 2}$, we know that ${\lim}_{x \rightarrow - 2} \frac{\left\mid x + 2 \right\mid}{x + 2}$ does not exist.

In fact, we saw that the function is simply $y = - 1$ when $x < - 2$ and is $y = 1$ when $x > - 2$.

Graphed is $\frac{\left\mid x + 2 \right\mid}{x + 2}$:

graph{abs(x+2)/(x+2) [-13.77, 8.73, -5.62, 5.63]}