Question

Lim x goes to 1 x^2-1/|x-1|
anyone can find this limit?

Answer 1

\[\lim_{x\to 1}\frac{x^2-1}{|x-1|}\]

Answer 2

seems unlikely since this is is a piecewise function
it is one thing if \(x>1\) and quite another if \(x<1\)

Answer 3

if \(x>1\) then \(|x-1|=x-1\)
when you factor and cancel you get
\[x+1\]

Answer 4

you get something else if \(x<1\) try it and see

Answer 5

I wonder if I should use left and right limit to do this

Answer 6

you have no choice but to do that, since \(|x-1|\) is a piecewise function which changes definition at \(x=1\)

Answer 7

lets take it step by step

Answer 8

if \(x>1\) then \(|x-1|=x-1\) right?

Answer 9

did i lose you there?

Answer 10

im here

Answer 11

if \(x>1\) then \(|x-1|=x-1\)so your function is
\[\frac{x^2-1}{x-1}=\frac{(x+1)(x-1)}{x-1}=x+1\]
making the limit as \(x\to 1^+=1+1=2\)

Answer 12

I see, now the left limit I guess

Answer 13

left limit is different because if \(x<1\) then \(|x-1|=1-x\)

Answer 14

so the limit doesnt exist right

Answer 15

no the two sided limit does not exist

Answer 16

i see

Answer 17

Thank you