Answer
$$\lim _{x \rightarrow a} \frac{(x+a)^{2}-4 x^{2}}{x-a} =-4a $$
Work Step by Step
Given $$\lim _{x \rightarrow a} \frac{(x+a)^{2}-4 x^{2}}{x-a} \ \ \ a \ \ \text{is constant}$$
let $$ f(x) = \frac{(x+a)^{2}-4 x^{2}}{x-a}$$
Since, we have
$$ f(a)= \frac{4 a ^2-4 a^{2}}{a-a}=\frac{0}{0}$$
So, we get
\begin{aligned}
L&=\lim _{x \rightarrow a} \frac{(x+a)^{2}-4 x^{2}}{x-a}\\
&=\lim _{x \rightarrow a} \frac{(x^2+2ax+a^2)-4 x^{2}}{x-a}\\
&=\lim _{x \rightarrow a} \frac{(-3x^2+2ax+a^2) }{x-a}\\
&=\lim _{x \rightarrow a} \frac{(x-a)(-3x-a) }{x-a}\\
&=\lim _{x \rightarrow a} (-3x-a) \\
&=(-3a-a)
\\
&=-4a
\end{aligned}
