Question

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.

Answer 1

Check below

Explanation:

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

`lim_(xto0)(|x-1|)/(x-1)=(|0-1|)/(0-1)=|-1|/(-1)=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->1^+` then `x>1` `<=>` `x-1>0` which means that the limit would be `lim_(xto1^+)(|x-1|)/(x-1)=lim_(xto1^+)(x-1)/(x-1)=1`

If `x` approaches `1` from the left then `x->1^(-)` which means `x<1` `<=>` `x-1<0` and

`lim_(xto1^(-))(|x-1|)/(x-1)=lim_(xto1^(-))(-(x-1))/(x-1)=-1`

Because `lim_(xto1^+)(|x-1|)/(x-1)!=lim_(xto1^(-))(|x-1|)/(x-1)`

`lim_(xto1)(|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->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`