Question

find the limit if it exists:
lim x->3= (x-3)/(x^2-9)

Answer 1

what i did was (x-3)/(x-3)(x+3)

Answer 2

it became 1/(x+3) then i plugged in the 3 into the x and got 1/6, this is wrong though.

Answer 3

no it is not. 1/6 is the right answer

Answer 4

hang on

Answer 5

1/6 is correct or verify the question.

Answer 6

yea ur right who told u ur wrong

Answer 7

oh wait i read the wrong answer in the back of my book. it is correct, thanks all!

Answer 8

i wish i could give everyone a medal :(

Answer 9

\[\lim_{x\to3}\frac{x-3}{x^2-9}=\lim_{x\to3}\frac{x-3}{(x+3)(x-3)}=\lim_{x\to3}\frac1{x+3}=1/6\]