Question

Solve $\underset{x\to -2}{lim}\frac{x^4-16}{x^3+8}$

Answer 1

$li{m}_{x\to -2}\frac{x^4-16}{x^3+8}$

$=li{m}_{x\to -2}\frac{(x^2+4)(x+2)(x-2)}{(x+2)(x^2+4-2x)}$

$=li{m}_{x\to -2}\frac{(x^2+4)(x-2)}{x^2+4-2x}$

$=\frac{8(-4)}{12}=\frac{-32}{12}$

$=\frac{-8}{3}$