Answer
$$\lim _{h \rightarrow 4} \frac{(h+2)^2-9h}{h-4}=3$$
Work Step by Step
Given $$\lim _{h \rightarrow 4} \frac{(h+2)^2-9h}{h-4}$$
let $$ f(h) = \frac{(h+2)^2-9h}{h-4}$$
Since, we have
$$ f(4)=\frac{36-36}{4-4}=\frac{0}{0}$$
So, transform algebraically and cancel
\begin{aligned}
L&=\lim _{h \rightarrow 4} \frac{(h+2)^2-9h}{h-4}\\
&= \lim _{h \rightarrow 4} \frac{h^2+4h+4-9h}{h-4}\\
&= \lim _{h \rightarrow 4} \frac{h^2-5h+4}{h-4}\\
&= \lim _{h \rightarrow 4} \frac{(h-4)(h-1)}{h-4}\\
&= \lim _{h \rightarrow 0}(h-1)\\
&=4-1\\
&= 3
\end{aligned}
