What is $\lim_{x\to 7^-} \frac{\left|x-7\right|}{x-7}\,$?

Question

$\displaystyle \lim_{x\to 7^-} \frac{\left|x-7\right|}{x-7} = $

Writing absolute value as:

$x-7 > 0$

$x > 7$

which means

$x - 7$ when $x > 7$

then:

$ -(x - 7) < 0$

$-x + 7 < 0$

$-x < - 7$

$x > 7$

which means

$-x + 7 $ when $x > 7$

So when $x > 7$ what equation should I use?

$x - 7$

or

$-x + 7 $

Answer 1

$$|x-7| = \begin{cases} x-7, & x\ge 7 \\ -(x-7), & x\lt 7\end{cases}$$

Your limit is from the left side (I can tell because you've written $x\to 7^-$), meaning that you only care what this limit approaches for values of $x\lt 7$. And in this region of the domain of your function $|x-7| = -(x-7)$. So

$$\lim_{x\to 7^-} \dfrac {|x-7|}{x-7} = \lim_{x\to 7^-} \dfrac {-(x-7)}{x-7} = \lim_{x\to 7^-} -1 = -1$$

Answer 2

The limit from the left means that $x<7$, that is, $x-7<0$, that is, $|x-7|=-(x-7)$.

Answer 3

Perhaps a graph would help here before trying to "prove" anything. For $f(x)=\frac{|x-7|}{x-7}$, we have the following graph for a sample viewing window:

As you can see from the graph of $f(x)$, when $x$ approaches $7$ from the right (i.e, $x\to7^+$), we have that $f(x)=1$; on the other hand, when $x$ approaches $7$ from the left (i.e., $x\to7^-$), we have that $f(x)=-1$. Dissecting what $f(x)=\frac{|x-7|}{x-7}$ actually means will make it easy to evaluate your limit. As Bye_World pointed out, $$ |x-7| = \begin{cases} x-7, & x\ge 7 \\ -(x-7), & x\lt 7\end{cases}. $$ Hence, we have the following: $$ \lim_{x\to 7^+}f(x) = \lim_{x\to 7^+}\frac{|x-7|}{x-7}=\lim_{x\to 7^+}\frac{x-7}{x-7}=1, $$ and $$ \lim_{x\to 7^-}f(x) = \lim_{x\to 7^-}\frac{|x-7|}{x-7}=\lim_{x\to 7^-}\frac{-(x-7)}{x-7}=-1. $$ The figure above confirms this. Also note that $\lim_{x\to 7}f(x)$ does not exist since $\lim_{x\to 7^+}f(x)=1\neq-1=\lim_{x\to7^-}f(x)$.