Question

evaluate the limit as approaches -2 of the function (x+2)/(x^3 + 8)

Answer 1

the form is 0/0 so we can use L'Hopitals rule

Answer 2

take derivatives of top and bottom and then we can plug in -2

Answer 3

you can also factor if you don't know L'Hopitals rule

Answer 4

One way is what rsvitale said. another way is to facto the bottom. You will get
\[\lim_{x \rightarrow -2}{x+2 \over (x+2)(x^2-2x+4)}=\lim_{x \rightarrow -2}{1 \over x^2-2x+4}={1 \over 12}\]

Answer 5

yep :)

Answer 6

:D

Answer 7

ok sweet thanks guys