Answer
$\lim\limits_{x \to -1}\dfrac{x^{3}+1}{x+1}=3$
Work Step by Step
$\lim\limits_{x \to -1}\dfrac{x^{3}+1}{x+1}$
Factor the numerator and simplify:
$\lim\limits_{x \to -1}\dfrac{x^{3}+1}{x+1}=\lim\limits_{x \to -1}\dfrac{(x+1)(x^{2}-x+1)}{x+1}=\lim\limits_{x \to -1}x^{2}-x+1$
Apply direct substitution to evaluate the limit:
$\lim\limits_{x \to -1}x^{2}-x+1=(-1)^{2}-(-1)+1=3$
