Answer
$$\lim _{h \rightarrow 0} \frac{\frac{1}{(h+2)^{2}}-\frac{1}{4}}{h}
=-\frac{1}{4} $$
Work Step by Step
Given $$\lim _{h \rightarrow 0} \frac{\frac{1}{(h+2)^{2}}-\frac{1}{4}}{h}$$
let $$ f(h) = \frac{\frac{1}{(h+2)^{2}}-\frac{1}{4}}{h}$$
Since, we have
$$ f(0)= \frac{\frac{1}{4}-\frac{1}{4}}{0}=\frac{0}{0}$$
So, transform algebraically and cancel
\begin{aligned}L&= \lim _{h \rightarrow 0} \frac{\frac{1}{(h+2)^{2}}-\frac{1}{4}}{h} \\
& = \lim _{h \rightarrow 0} \frac{\frac{4-(h+2)^{2}}{4(h+2)^{2}}}{h}\\
&= \lim _{h \rightarrow 0} \frac{\frac{4-\left(h^{2}+4 h+4\right)}{4(h+2)^{2}}}{h}\\
&= \lim _{h \rightarrow 0} \frac{4-h^{2}-4 h-4}{4(h+2)^{2} h}\\
&= \lim _{h \rightarrow 0} \frac{-h^{2}-4 h}{4 h(h+2)^{2}}\\
&= \lim _{h \rightarrow 0} \frac{-h(h+4)}{4 h(h+2)^{2}}\\
&=\lim _{h \rightarrow 0} \frac{-(h+4)}{4(h+2)^{2}} \\ &=\frac{-(0+4)}{4(0+2)^{2}} \\ &=\frac{-(4)}{4 (2)^{2}} \\
&=\frac{-(4)}{16} \\
&=-\frac{1}{4} \\
\end{aligned}
