## Calculus, 10th Edition (Anton)

-$\frac{4}{5}$
x² + 6x + 5 = (x + 1)(x + 5) and x² - 3x - 4 = (x + 1)(x - 4), then: $\frac{x²+6x+5}{x²-3x-4}$ = $\frac{x+5}{x-4}$. Now we have: $\lim\limits_{x \to -1}\frac{x²+6x+5}{x²-3x-4}$ = $\lim\limits_{x \to -1}\frac{x+5}{x-4}$ = $\frac{-1+5}{-1-4}$ = -$\frac{4}{5}$