# What is the limit of |x-1|/(x-1) as x approaches 0?

## a pretty basic question i assume, but i'm very confused. my textbook just says that it does have a limit (that was the original question) but i can't get to the same answer. here's something i've come up with: |x-1|=x-1, if x>1 and -(x-1) if x<1 the limit of |x-1|/(x-1)as x->0 from the left would be -(x-1)/(x-1)=-1 the limit of |x-1|/(x-1) as x-> from the right would be (x-1)/(x-1)=1 which then says that the limit of |x-1|/(x-1) as x->0 doesn't exist. so i'm a bit lost. thanks in advantage.

Jun 21, 2018

Check below

#### Explanation:

$| x - 1 |$ is not affected when $x$ is near $0$, it is affected when $x$ is approaching $1$

${\lim}_{x \to 0} \frac{| x - 1 |}{x - 1} = \frac{| 0 - 1 |}{0 - 1} = | - 1 \frac{|}{- 1} = \frac{1}{- 1} = - 1$

In the case that $x$ approaches $1$ we'll need to determine if it approaches $1$ from the left or right because if $x \to {1}^{+}$ then $x > 1$ $\iff$ $x - 1 > 0$ which means that the limit would be ${\lim}_{x \to {1}^{+}} \frac{| x - 1 |}{x - 1} = {\lim}_{x \to {1}^{+}} \frac{x - 1}{x - 1} = 1$

If $x$ approaches $1$ from the left then $x \to {1}^{-}$ which means $x < 1$ $\iff$ $x - 1 < 0$ and

${\lim}_{x \to {1}^{-}} \frac{| x - 1 |}{x - 1} = {\lim}_{x \to {1}^{-}} \frac{- \left(x - 1\right)}{x - 1} = - 1$

Because ${\lim}_{x \to {1}^{+}} \frac{| x - 1 |}{x - 1} \ne {\lim}_{x \to {1}^{-}} \frac{| x - 1 |}{x - 1}$

${\lim}_{x \to 1} \frac{| x - 1 |}{x - 1}$ does not exist.

Note: When we say "near" in mathematics we are referring to an infinitesimal small region, $0$ is not near $1$ so taking the sided limits when $x \to 0$ does not mean that the function will have the same behaviour as if $x$ was approaching $1$

graph{|x-1| [-2.03, 3.446, -1.114, 1.622]}
here is a graph of $y = | x - 1 |$ you can see that the sign changes near $1$ and not $0$