## Ephilo 2 years ago Find the limit of x^2-4/x^3+8 as x approaches -2 ?

I am assuming you have $\frac{ ( x ^{2}-4) }{ (x^{3}+8) }$ in which case you will need to factor the top and the bottom (difference of squares for the top and sum of cubes for the bottom) which will get you $lim_{ X \rightarrow -2} \frac{ ( x-2)(x+2) }{ (x+2)(x^2-2x+4) }$ nd then you can factor the x+2 out and plug in -2 for the answer