Answer
$\frac{1}{6}$
Work Step by Step
$\lim\limits_{h \to 0}\frac{\sqrt {9+h}-3}{h}=\frac{\sqrt {9+(0)}-3}{(0)}=\frac{0}{0}=$indet
so we have to do algebra
$\lim\limits_{h \to 0}\frac{\sqrt {9+h}-3}{h}=\lim\limits_{h \to 0}\frac{\sqrt {9+h}-3}{h}*\frac{\sqrt {9+h}+3}{\sqrt {9+h}+3}=\lim\limits_{h \to 0}\frac{9+h-9}{(h)(\sqrt {9+h}+3)}=\lim\limits_{h \to 0}\frac{1}{\sqrt {9+h}+3}=\frac{1}{\sqrt {9+(0)}+3}=\frac{1}{6}$
