Single Variable Calculus, Chapter 2, 2.3, Section 2.3, Problem 17

Determine the $\displaystyle\lim \limits_{h \to 0} \frac{(4 + h)^2 - 16}{h}$, if it exists.


$
\begin{equation}
\begin{aligned}
& \lim \limits_{h \to 0} \frac{(4 + h)^2 - 16}{h} = \frac{h^2 + 8h + \cancel{16} - \cancel{16}}{h}
&& \text{ Expand the equation and combine like terms. }\\
\\
& \lim \limits_{h \to 0} \frac{h^2 + 8h}{h} = \lim \limits_{h \to 0} \frac{\cancel{h} (h + 8)}{\cancel{h}}
&& \text{ Get the factor and cancel out like terms. }\\
\\
& \lim \limits_{h \to 0} h + 8 = 0 + 8
&& \text{ Substitute value of $h$ and simplify}\\
\\
& \fbox{$\lim \limits_{h \to 0} \displaystyle \frac{(4 + h)^2 - 16}{h} = 8 $}


\end{aligned}
\end{equation}
$