## Calculus: Early Transcendentals (2nd Edition)

The answer is $$\lim_{x\to -1}\frac{x^4+x^3+2x+2}{x+1}=1.$$
We will evaluate this limit using L'Hopital's rule. $\text{LR}$ will stand for "Apply L'Hopital's rule" $$\lim_{x\to -1}\frac{x^4+x^3+2x+2}{x+1}=\left[\frac{0}{0}\right][\text{LR}]=\lim_{x\to -1}\frac{(x^4+x^3+2x+2)'}{(x+1)'}=\lim_{x\to -1}\frac{4x^3+3x^2+2}{1}=\left[\frac{4\cdot(-1)^3+3\cdot(-1)^2+2}{1}\right] =1.$$