Question

lim x approaches 1 |(x-1)|/(x-1)

Answer 1

What did you get when you tried to evaluate this?

Answer 2

0? i plugged in 1 but i dont think thats correct..

Answer 3

See first convert modulus into function at different intervals

|x|=x when x>0
|x|=-x when x<0

Write this for |x-1|

Answer 4

don't be hoodwinked by this

Answer 5

\[\frac{|x-1|}{x-1}\] is only ever one of two numbers
if \(x-1>0\) then it is \(1\) and if \(x-1<0\) then it is \(-1\)
try a few numbers and you will see it instantly

Answer 6

so i just plug in numbers greater and less than 1?

Answer 7

lets go slow

Answer 8

what i meant was, if you plug in any number larger than 1 for x, you will get 1

Answer 9

\[\frac{|3-1|}{3-1}=\frac{2}{2}=1\] for example

Answer 10

so for any number larger than 1, this expression is just the number 1

Answer 11

and for any number less than one, this expression is just the number \(-1\)
for example, if \(x=0\) you get \[\frac{|0-1|}{0-1}=\frac{1}{-1}=-1\]

Answer 12

if x - 1 > and < aren't equal, does that mean the limit doesn't exist?

Answer 13

in other words \[\frac{|x-1|}{x-1} = \left\{\begin{array}{rcc}- 1& \text{if} & x <1 \\
1& \text{if} & x>1
\end{array}
\right.
\]

Answer 14

and yes, since the limit from the right is 1 and from the left is -1, that means the limit does not exist, like you said

Answer 15

ok thank you i understand

Answer 16

yw