## Calculus (3rd Edition)

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