5) Ta có
$\underset{x \to -1}{\lim} \dfrac{2x^3 + 3x + 5}{x^3 + 3x^2+x-1} = \underset{x \to -1}{\lim} \dfrac{(x+1)(2x^2 - 2x +5)}{(x+1)(x^2 + 2x-1)}$
$= \underset{x \to -1}{\lim} \dfrac{2x^2 - 2x + 5}{x^2 + 2x - 1}$
$= \dfrac{2(-1)^2 - 2(-1) + 5}{(-1)^2 + 2(-1) - 1} = -\dfrac{9}{2}$