Question

Hey, can y'all check this:
\[\lim_{ {x->-2^{-}}} \frac{|x^2-x-6|}{(x^2-4x+4)}\]

The book is saying the answer is \(+\infty \), but I am getting 0.

Answer 1

Did you factor them both or what?

Answer 2

I found for which x the abs val is positive, -2 is the branch pt, then because the fn exists at x=-2 I plugged in the corresponding abs val. The issue is that it is a cusp, but I just can't remember how to get infinity.

Answer 3

and, I just did it with factoring and I'm still getting 0.

Answer 4

not sure if I'm just stupid, or the book is wrong

Answer 5

\[
\lim_{ {x->2^{-}}} \frac{|x^2-x-6|}{(x^2-4x+4)}= \frac{|2^2-2-6|}{(2^2-4(2)+4)}=4/0=\infty
\]

Answer 6

@eliassaab Sorry, missed the -. It's -2

Answer 7

It is \(+\infty\) since the numerator and the denominator are positive near 2

Answer 8

what the f#%ck

Answer 9

@eliassaab Can you elaborate on that last bit?

Answer 10

It is \( 2^{- }\), which means we approach 2 from the left.

Answer 11

(-2)^2 -4(-2) + 4 = 4+8+4 = not zero

Answer 12

no its \((-2)^{-}\)

Answer 13

right, but the limit is 0/16 if you just plug and chug

Answer 14

If it is -2 than the answer is zero. if 2, it is +infinity

Answer 15

right, I agree it's zero, I need to prove the book wrong

Answer 16

graphically confirmed that 2^- = +infinity
-2^- = 0

Answer 17

the kink is located at -2

Answer 18

http://www.wolframalpha.com/input/?i=graph+%7Cx%5E2-x-6%7C%2F%28x%5E2-4x%2B4%29

Answer 19

but I thought that limit DNE at kinks

Answer 20

I know there is a cusp, but I just don't know how to show that without being allowed a graphing calculator

Answer 21

@satellite73

Answer 22

but anyways it's -2 not 2

Answer 23

\[ |x^2-x-6| = \left\{\begin{array}{rcc}
(x-3)(x+2) & \text{if} & (x-3)(x+2) \geq 0 \\
-(x-3)(x+2)& \text{if} & (x-3)(x+2) <0
\end{array}
\right.\]

Answer 24

since you are going to \(-2\) from below, both \((x-3)\) and \((x+2)\) are negative

Answer 25

and so \(|x^2-x-6|=(x-3)(x+2)\)

Answer 26

oops that last part is wrong

Answer 27

you get
\[\frac{(x-3)(x+2)}{(x-2)^2}\]

Answer 28

wait a sec
why can't you just plug in \(-2\)? the denominator will not be zero
the numerator will
you get 0

Answer 29

oh i see, book says \(\infty\)
either there is a typo in the problem or in the answer
you just get zero

Answer 30

thank you @satellite73

Answer 31

lol that's what i get for coming in late
question was already answered

Answer 32

we just needed another person to look at

Answer 33

at the problem

Answer 34

Thanks guys, I thought there was a type-o. You have confirmed it! I really appreciate everyone's help!