## Calculus: Early Transcendentals 8th Edition

$\lim\limits_{h\to0}\frac{(3+h)^{-1}-3^{-1}}{h}=\frac{-1}{9}$
$\lim\limits_{h\to0}\frac{(3+h)^{-1}-3^{-1}}{h}$ $=\lim\limits_{h\to0}\frac{\frac{1}{3+h}-\frac{1}{3}}{h}$ $=\lim\limits_{h\to0}\frac{\frac{3-(3+h)}{3(3+h)}}{h}$ $=\lim\limits_{h\to0}\frac{\frac{-h}{3(3+h)}}{h}$ $=\lim\limits_{h\to0}\frac{-h}{3h(3+h)}$ $=\lim\limits_{h\to0}\frac{-1}{3(3+h)}$ (divide both numerator by denominator by $h$) $=\frac{-1}{3(3+0)}$ $=\frac{-1}{9}$