## Calculus, 10th Edition (Anton)

Published by Wiley

# Chapter 1 - Limits and Continuity - 1.2 Computing Limits - Exercises Set 1.2 - Page 69: 9

#### Answer

-$\frac{4}{5}$

#### Work Step by Step

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: $\lim\limits_{x \to -1}\frac{x²+6x+5}{x²-3x-4}$ = $\lim\limits_{x \to -1}\frac{x+5}{x-4}$ = $\frac{-1+5}{-1-4}$ = -$\frac{4}{5}$

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