Evaluate the Limit Without L'Hopital Rule Let $\lim_{x\to a}\frac{a^4-(x^2-x\left | x \right |-a^2)^2}{x-a}=L$, find the value of $\lim_{x\to a}\frac{x(x^2-x\left | x \right |-a^2)^2-a^4\left | a \right |}{x-a}$ for $a\neq0$. 
Using L'Hopital rule I found that the answer is $a^4-aL$. My question is how to solve this problem without using L'Hopital rule.
Here's my attempt using L'Hopital.
$$
\begin{aligned}
\lim_{x\to a}\frac{a^4-(x^2-x\left | x \right |-a^2)^2}{x-a}&=L\\
\lim_{x\to a}\frac{d}{dx}(x^2-x\left | x \right |-a^2)^2&=-L...(1)\\
\end{aligned}
$$
Let the numerator equal to zero.
$$
(x^2-x\left | x \right |-a^2)^2=a^4...(2)
$$
Ergo
$$\begin{aligned}
\lim_{x\to a}\frac{(x^2-x\left | x \right |-a^2)^2-a^4\left | a \right |}{x-a}&=\lim_{x\to a}(x^2-x\left | x \right |-a^2)^2+x\cdot\frac{d}{dx}(x^2-x\left | x \right |-a^2)^2\\
&=a^4-aL
\end{aligned}
$$

Hint: The numerator can be written as $$-x \left(  \left| x \right| -x \right)  \left( x \left| x \right| +2\,
{a}^{2}-{x}^{2} \right) 
$$ and now distinguish the cases $$x\geq 0$$ or $$x<0$$
