Question

lim x-->8 (x-8/x^2-64)

Answer 1

Can you factor the denominator into anything?
Hint: difference of squares

Answer 2

?

Answer 3

\[\lim_{x \rightarrow 8} \frac {x-8}{x^2 - 64} = \lim_{x \rightarrow 8} \frac {x-8}{(x-8)(x+8)}\]

Answer 4

oh yeah

Answer 5

Your answer should be 1/16. Did you get it?

Answer 6

no

Answer 7

I know the (x-8) cancels
which leaves 1/ x+8

Answer 8

\[\lim_{x \rightarrow 8} \frac {1}{(x+8)} = \frac {1}{16}\]

Answer 9

Yeah, so just plug in 8.

Answer 10

That's what x approaches in the limit.

Answer 11

got it! thank you!

Answer 12

Alrighty, no problem :)