Answer
$$\lim _{h \rightarrow 0} \frac{2(a+h)^{2}-2 a^{2}}{h}=4a $$
Work Step by Step
Given $$\lim _{h \rightarrow 0} \frac{2(a+h)^{2}-2 a^{2}}{h}, \ \ \ a \ \ \text{is constant}$$
let $$ f(h) = \frac{2(a+h)^{2}-2 a^{2}}{h}$$
Since, we have
$$ f(0)= \frac{2 a ^2-2 a^{2}}{0}=\frac{0}{0}$$
So, we get
\begin{aligned}
L&=\lim _{h \rightarrow 0} \frac{2(a+h)^{2}-2 a^{2}}{h}\\
&=\lim _{h \rightarrow 0} \frac{(2 a^2+4ah+2h^2) -2 a^{2}}{h}\\
&=\lim _{h \rightarrow 0} \frac{( 4ah+2h^2) }{h}\\
&=\lim _{h \rightarrow 0} \frac{h( 4a +2h ) }{h}\\
&=\lim _{h \rightarrow 0} ( 4a +2h )\\
&=( 4a +0 )\\
&=4a
\end{aligned}
