How to compute $\lim\limits_{x \to a} \frac{\sqrt{2 a^{3} x-x^{4}}-a \sqrt[3]{a^{2} x}}{a-\sqrt[4]{a x^{3}}}$? 
How to compute $\lim\limits_{x \to a} \frac{\sqrt{2 a^{3} x-x^{4}}-a \sqrt[3]{a^{2} x}}{a-\sqrt[4]{a x^{3}}}$?

My process:
\begin{align}
\lim\limits_{x \to a} \frac{\sqrt{2 a^{3} x-x^{4}}-a \sqrt[3]{a^{2} x}}{a-\sqrt[4]{a x^{3}}}=\frac{\sqrt{2 a^{3} a-a^{4}}-a \sqrt[3]{a^{2} a}}{a-\sqrt[4]{a a^{3}}}=\frac{\sqrt{a^4}-a\cdot \sqrt[3]{a^3}}{a-\sqrt[4]{a^4}}=\frac{a^2-a\cdot a}{a-a}=\frac{a^2-a^2}{a-a}
\end{align}
which would be undefined, but Wolfram Alpha calculated this:

$\lim_{x\to a} \, \frac{\sqrt{2 a^3 x-x^4}-a \sqrt[3]{a^2 x}}{a-\sqrt[4]{a x^3}}=
\begin{array}{cc}
 \{ & 
\begin{array}{cc}
 \frac{16 a}{9} & a\geq 0 \\
 0 & (\text{otherwise}) \\
\end{array}
 \\
\end{array}$

What have I done wrong? Is there a way to calculate the limit with L'Hospitals Rule?

Similar to @user's answer, starting with $x=a y$, we have
$$A=\frac{\sqrt{2 a^{3} x-x^{4}}-a \sqrt[3]{a^{2} x}}{a-\sqrt[4]{a x^{3}}}=a\frac{ \sqrt[3]{y}-\sqrt{y \left(2-y^3\right)}}{y^{3/4}-1}$$ Now, let $y=1+t$ to get
$$A=a\frac{\sqrt[3]{1+t}-\sqrt{1-2 t-6 t^2-4 t^3-t^4}}{(1+t)^{3/4}-1}$$ Now, using Taylor series around $t=0$ or binomial expansion
$$A=a \left(\frac{16}{9}+\frac{128 }{27}t+O\left(t^2\right) \right)$$ which shows the limit and also how it is approached.
