Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 2 - Limits - Chapter Review Exercises - Page 94: 29


$$-\frac{1}{ 2 }$$

Work Step by Step

By substituting, we get the intermediate form $\frac{0}{0}$. So, by factorizing the denominator, we have $$ \lim _{z \rightarrow -3 } \frac{z+3}{z^2+4z+3}=\lim _{z \rightarrow -3 } \frac{z+3}{(z+3)(z+1)}\\ =\lim _{z \rightarrow -3 } \frac{1}{ z+1 }=\frac{1}{ -3+1 }=-\frac{1}{ 2 }. $$
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