## Calculus: Early Transcendentals 8th Edition

$\lim\limits_{h\to 0}\frac{\sqrt{9+h}-3}{h}=\frac{1}{6}$
$A=\lim\limits_{h\to 0}\frac{\sqrt{9+h}-3}{h}$ Multiply both numerator and denominator by $\sqrt{9+h}+3$, we have $A=\lim\limits_{h\to 0}\frac{(9+h)-9}{h(\sqrt{9+h}+3)}$ (We see that ($\sqrt{9+h}-3)(\sqrt{9+h}+3)=(9+h)-9$ since $(a-b)(a+b)=a^2-b^2$) $A=\lim\limits_{h\to 0}\frac{h}{h(\sqrt{9+h}+3)}$ $A=\lim\limits_{h\to 0}\frac{1}{\sqrt{9+h}+3}$ (divide numerator and denominator by $h$) $A=\frac{1}{\sqrt{9+0}+3}$ $A=\frac{1}{6}$