Answer
$\lim_{h\to0}\dfrac{(3+h)^{-1}-3^{-1}}{h}=-\dfrac{1}{9}$
Work Step by Step
$\lim_{h\to0}\dfrac{(3+h)^{-1}-3^{-1}}{h}$
Rewrite $\dfrac{(3+h)^{-1}-3^{-1}}{h}$ as $\dfrac{\dfrac{1}{3+h}-\dfrac{1}{3}}{h}$
$\lim_{h\to0}\dfrac{(3+h)^{-1}-3^{-1}}{h}=\lim_{h\to0}\dfrac{\dfrac{1}{3+h}-\dfrac{1}{3}}{h}=...$
Try to evaluate the limit applying direct substitution:
$\lim_{h\to0}\dfrac{\dfrac{1}{3+h}-\dfrac{1}{3}}{h}=\dfrac{\dfrac{1}{3+0}-\dfrac{1}{3}}{0}=\dfrac{0}{0}$ Indeterminate form
The limit could not be evaluated using direct substitution. Evaluate the subtraction of fractions in the numerator and simplify:
$\lim_{h\to0}\dfrac{\dfrac{1}{3+h}-\dfrac{1}{3}}{h}=\lim_{h\to0}\dfrac{\dfrac{3-(3+h)}{3(3+h)}}{h}=...$
$...=\lim_{h\to0}\dfrac{3-(3+h)}{3h(3+h)}=\lim_{h\to0}\dfrac{-h}{3h(3+h)}=...$
$...=\lim_{h\to0}\dfrac{-1}{3(3+h)}=...$
Try to evaluate the limit applying direct substitution again:
$\lim_{h\to0}\dfrac{-1}{3(3+h)}=\dfrac{-1}{3(3+0)}=-\dfrac{1}{9}$
