Single Variable Calculus, Chapter 2, 2.3, Section 2.3, Problem 28

Determine the $\displaystyle\lim \limits_{h \to 0} \frac{(3 + h)^{-1} - 3^{-1}}{h}$, if it exists.


$
\begin{equation}
\begin{aligned}
& \lim \limits_{h \to 0} \frac{(3 + h)^{-1} - 3^{-1}}{h}
= \lim \limits_{h \to 0} \frac{\frac{1}{3 + h} - \frac{1}{3}}{h}
&& \text{ Get the reciprocal of the numerator to eliminate negative exponents.}\\
\\
& \lim \limits_{h \to 0} \frac{\frac{1}{3 + h} - \frac{1}{3}}{h}
= \lim \limits_{h \to 0} \frac{\frac{3-(3+h)}{3(3+h)}}{h} = \lim \limits_{h \to 0} \frac{3-(3+h)}{3h(3+h)}
&& \text{ Get the LCD and simplify}\\
\\
& \lim \limits_{h \to 0} \frac{-1}{9 + 3h}
= \frac{-1}{9 + 3(0)}
= \frac{-1}{9}
&& \text{ Substitute value of $x$ then simplify}\\
\\
& \fbox{$ \lim \limits_{h \to 0} \displaystyle \frac{( 3 + h)^{-1} - 3^{-1}}{h} = \frac{-1}{9} $}

\end{aligned}
\end{equation}
$