Question

Limit as x->1^- (x-1)/(absolute value(x^2-1))

How is that -1/2 instead of 1/2....

Answer 1

\[\lim_{x\rightarrow 1}(\frac{x-1}{|x^2-1|})\]
is this the question?

Answer 2

1^-

Answer 3

Ah, the issue is the jump/hole at x=1
from the left, at 1 y=-1/2
while, from the right at 1 y=1/2

Answer 4

http://www4b.wolframalpha.com/Calculate/MSP/MSP17091a0hfg6961dh04e50000282g65b0gfc7f932?MSPStoreType=image/gif&s=3&w=413&h=190&cdf=RangeControl

Answer 5

Ahhhhhhh its x-1/-(x^2-1) because x<1

Answer 6

indeed