Question

Calcule o limite abaixo:

Answer 1

$\lim\limits_{x\to -1}\dfrac{x^2-2x}{x^2-4x+4}=\dfrac{(-1)^2-2(-1)}{(-1)^2-4(-1)+4}=\dfrac{1+2}{1+4+4}=\dfrac{3}{9}=\dfrac{1}{3}$