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.

1 Answer
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_(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#