<s>[INST] 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?
 [/INST] We have that by $y=x/a\to 1$
$$\frac{\sqrt{2 a^{3} x-x^{4}}-a \sqrt[3]{a^{2} x}}{a-\sqrt[4]{a x^{3}}}=a \frac{\sqrt{2 y-y^{4}}-\sqrt[3]{y}}{1-\sqrt[4]{y^3}}=$$
$$= -a \frac{y-1}{\sqrt[4]{y^3}-1}\left( \frac{\sqrt{2 y-y^{4}}-1}{y-1}- \frac{\sqrt[3]{y}-1}{y-1}\right)=f(y)$$
then use the definition of derivative to obtain
$$f(y)\to -a\cdot \frac43\left(-1-\frac13\right)=\frac{16a}9$$
</s>