Answer
The solution is
$$\lim_{x\to-1}\frac{x^3-x^2-5x-3}{x^4+2x^3-x^2-4x-2}=4.$$
Work Step by Step
To solve this limit we will use L'Hopital's rule. "LR" will stand for "Apply L'Hopital's rule."
$$\lim_{x\to-1}\frac{x^3-x^2-5x-3}{x^4+2x^3-x^2-4x-2}=\left[\frac{0}{0}\right][\text{LR}]=\lim_{x\to-1}\frac{(x^3-x^2-5x-3)'}{(x^4+2x^3-x^2-4x-2)'}=\lim_{x\to-1}\frac{3x^2-2x-5}{4x^3+6x^2-2x-4}=\left[\frac{3(-1)^2-2(-1)-5}{4(-1)^3+6(-1)^2-2(-1)-4}\right]=\left[\frac{0}{0}\right][\text{LR}]=\lim_{x\to-1}\frac{(3x^2-2x-5)'}{(4x^3+6x^2-2x-4)'}=\lim_{x\to-1}\frac{6x-2}{12x^2+12x-2}=\frac{6(-1)-2}{12(-1)^2+12(-1)-2}=4$$
