Question

Find the limit (if it exists)...

Answer 1

\[\lim_{x \rightarrow 2}\frac{ x-2 }{x^2+4x-12 }\]

Answer 2

I know the answer is 1/8 but I am not sure how to get it.. help is much appreciated

Answer 3

Well, if we just plug in 2, we get an indeterminant form...

so we need to do some simplification first..looks like the denominator can be factored doesnt it :)

Answer 4

Yeah... like i keep getting 1/0 but I know thats not the answer, so its not as simple as just plugging it in right?

Answer 5

try factoring the denominator

Answer 6

Correct, and in general (assuming this is precalc maybe?) whenever you get an indeterminant form of (0/0) you will need to simplify to continue. So lets factor the bottom of the fraction and see where that gets us

Answer 7

X^2+6x-2x-12=0
x(x+6)-2(x+6)=0
(x+6)(x-2)=0

Answer 8

is that right?

Answer 9

Correct! so...

Answer 10

Now we have

\[\large \frac{(x - 2)}{(x + 6)(x - 2)}\]

notice anything?

Answer 11

(2+6)(2-2)
8+0=8?
1/8?

Answer 12

Am i adding the bottom or multiplying?

Answer 13

we have an x - 2 on both ends of the fraction so...

\[\large \frac{\cancel{(x - 2)}}{(x + 6)\cancel{(x - 2)}}\]

which leaves us with

\[\large \frac{1}{x + 6}\]

Answer 14

ohhhhhhhh!! hehee okay i see!! yay thank you so much!!

Answer 15

Haha there we go :) and anytime! Need anything else feel free to tag or message me :)

Answer 16

Okay, thank you so much! I am about to post one more question in a few minutes that I am having trouble with, I would love the help if you're able to. Thanks again!

Answer 17

Of course :)