Answer
$$\lim _{h \rightarrow 0} \frac{(3 a+h)^{2}-9 a^{2}}{h}=6a $$
Work Step by Step
Given $$\lim _{h \rightarrow 0} \frac{(3 a+h)^{2}-9 a^{2}}{h}, \ \ \ a \ \ \text{is constant}$$
let $$ f(h) = \frac{(3 a+h)^{2}-9 a^{2}}{h}$$
Since, we have
$$ f(0)= \frac{9 a ^2-9 a^{2}}{0}=\frac{0}{0}$$
So, we get
\begin{aligned}
L&=\lim _{h \rightarrow 0} \frac{(3 a+h)^{2}-9 a^{2}}{h}\\
&=\lim _{h \rightarrow 0} \frac{(9 a^2+6ah+h^2) -9 a^{2}}{h}\\
&=\lim _{h \rightarrow 0} \frac{( 6ah+h^2) }{h}\\
&=\lim _{h \rightarrow 0} \frac{h( 6a +h ) }{h}\\
&=\lim _{h \rightarrow 0} ( 6a +h )\\
&=( 6a +0 )\\
&=6a
\end{aligned}
