Question

Can someone please help find the limit of the absolute value of x^2-1 divided by x^2-6x+5 as x approaches 1 from the left

Answer 1

i would factor and cancel first

Answer 2

Can you cancel though? Because it is absolute value

Answer 3

don't see why not

Answer 4

That's just what my professor told us

Answer 5

you are taking a limit in any case, this thing is undefined at 1, but x cannot be 1

Answer 6

Right

Answer 7

http://www.answers.com/Q/What_is_the_limit_as_x_approaches_1_when_x-1_divided_by_absolute_value_of_x-1?#slide=1

Answer 8

might help?

Answer 9

-1/2???

Answer 10

go ahead and factor and cancel

Answer 11

the limit cannot be negative, that is for sure, as this is never negative

Answer 12

Hmmmm

Answer 13

it matters not that they are inside an absolute value, you can still factor and cancel since x is not 1

Answer 14

people say you can't factor and cancel absolute value but I've been doing it for years and never got one of those problems wrong. anyway, no one tells you why not

Answer 15

So your with absolute value x+1 over (x-5)

Answer 16

yes,
now replace x by 1

Answer 17

Yeah I got absolute value of 2/-4 so -1/2

Answer 18

absolute value is still there !

Answer 19

So 1/2?

Answer 20

yes

Answer 21

So it's applied to the whole problem not just the top?

Answer 22

i am not sure what you are asking

Answer 23

so it started absolute value of x^2-1 but it didn't have absolute value on the bottom the absolute value applies to the bottom also?

Answer 24

ooh i didn't see that part, lets check

Answer 25

yeah now it is a totally different story

Answer 26

That's why I wasn't sure if you could cancel because the bottom doesn't have absolute value and the top does but idk what else you could do algebrically

Answer 27

so lets back up and go slow

Answer 28

Ok

Answer 29

first off, is \(x>1\) or is \(x<1\) ?

Answer 30

X<1 because it's the left of 1? Right?

Answer 31

yes exactly
now we can remove the absolute value sign in the numerator, because if\(x<1\) then \(x^2-1<0\) right ?

Answer 32

Yes

Answer 33

in other words \(x^2-1\) is negative
that means \[|x^2-1|=1-x^2\] (only since \(x<1\) )

Answer 34

so now the absolute value signs are gone, and you can compute \[\lim{x\to -^1}\frac{1-x^2}{(x-5)(x-5)}\] by cancelling

Answer 35

Ok that makes sense!

Answer 36

typo there, i meant \[\lim_{x\to 1^-}\frac{1-x^2}{(x-5)(x-5)}\]

Answer 37

dang another typo!!

Answer 38

Lol it's ok!

Answer 39

\[\lim_{x\to 1^-}\frac{1-x^2}{(x-5)(x-1)}\]

Answer 40

now, i believe, the answer will in fact be \(-\frac{1}{2}\)
check it

Answer 41

Ok

Answer 42

Yes that is what I got thank you so much!!!

Answer 43

yw