## Precalculus (6th Edition) Blitzer

$\underset{x\to -2}{\mathop{\lim }}\,\left| x+2 \right|=0$, because as x-approaches $-2$, the value of $f\left( x \right)$ gets closer to 0.
Consider the provided function, $f\left( x \right)=\left| x+2 \right|$. To plot the graph of the function $f\left( x \right)=\left| x+2 \right|$ substitute different values of x in the equation $f\left( x \right)=\left| x+2 \right|$ to get different values of $f\left( x \right)$. Now, consider the provided limit, $\underset{x\to -2}{\mathop{\lim }}\,\left| x+2 \right|$. Consider the graph of the function $f\left( x \right)=\left| x+2 \right|$. To find $\underset{x\to -2}{\mathop{\lim }}\,\left| x+2 \right|$, examine the portion of the graph near $x=-2$. As x gets closer to $-2$, the value of $f\left( x \right)$ gets closer to the y-coordinate of 0. This point $\left( -2,0 \right)$ is as shown in the above graph. The point $\left( -2,0 \right)$ has a y-coordinate of 0. Thus, $\underset{x\to -2}{\mathop{\lim }}\,\left| x+2 \right|=0$. Hence the value of $\underset{x\to -2}{\mathop{\lim }}\,\left| x+2 \right|$ is 0.