Question

lim x tendendo a -2 de 8+x³/4-x²

Answer 1

$\lim_{x \to -2} \frac{8-x^3}{4-x^2} = \lim_{x \to -2} \frac{2^3+x^3}{2^2-x^2} \\ =\lim_{x \to -2} \frac{(2+x)(4-2x+x^2)}{(2-x)(2+x)}= \lim_{x \to -2} \frac{(x^2-2x+4)}{(2-x)}= \frac{12}{4}=3$