Question

lim approaches x==> -2

Answer 1

\[\sqrt[3]{\frac{ x+2 }{ x^{3}+8 }}\]

Answer 2

if you plug in x = -2 you get the result 0/0
so I would try using L'hopital's rule

Answer 3

ok

Answer 4

\[\sqrt[3]\frac{x+2}{x^3+8}=\sqrt[3]\frac{x+2}{(x+2)(x^2-2x+4)}=\frac{1}{\sqrt[3]{x^2-2x+4}}\]

Answer 5

thats the way

Answer 6

thanks :D

Answer 7

just plug in x = -2 and you have it