Question

solve for 16x^2/3 divide 4-x^4/3 where x is approaches by -8...

Answer 1

\[\lim_{x\rightarrow -8}\frac{16x^{\frac{2}{3}}}{4-x^{\frac{4}{3}}}\]?

Answer 2

i need an answer can you?

Answer 3

what happens when you replace x by -8?

Answer 4

i think that is all you have to do because this is not of the form 0/0

Answer 5

\[16(-8)^{\frac{2}{3}}=16\times 4=64\]

Answer 6

\[4-(-8)^{\frac{4}{3}}=4-16=-4\]

Answer 7

and
\[\frac{64}{-4}=-16\]

Answer 8

so if the questions is right it is a matter of evaluating at -8