Answer
$$
\begin{aligned}
\lim _{x \rightarrow a} \frac{\sqrt{2 a^{3} x-x^{4}}-a \sqrt[3]{a a x}}{a-\sqrt[4]{a x^{3}}}=\frac{16}{9} a
\end{aligned}
$$
Work Step by Step
$$
\begin{aligned}
&\lim _{x \rightarrow a} \frac{\sqrt{2 a^{3} x-x^{4}}-a \sqrt[3]{a a x}}{a-\sqrt[4]{a x^{3}}}=\\
& \,\ [ \text{form is} \,\, 0 / 0\,\, \text {and by using L'Hôpital's rule we have} ]\\
&= \lim _{x \rightarrow a} \frac{\frac{1}{2}\left(2 a^{3} x-x^{4}\right)^{-1 / 2}\left(2 a^{3}-4 x^{3}\right)-a\left(\frac{1}{3}\right)(a a x)^{-2 / 3} a^{2}}{-\frac{1}{4}\left(a x^{3}\right)^{-3 / 4}\left(3 a x^{2}\right)}\\
& =\frac{\frac{1}{2}\left(2 a^{3} a-a^{4}\right)^{-1 / 2}\left(2 a^{3}-4 a^{3}\right)-\frac{1}{3} a^{3}\left(a^{2} a\right)^{-2 / 3}}{-\frac{1}{4}\left(a a^{3}\right)^{-3 / 4} \left(3 a a^{2}\right)}\\
&=\frac{\left(a^{4}\right)^{-1 / 2}\left(-a^{3}\right)-\frac{1}{3} a^{3}\left(a^{3}\right)^{-2 / 3}}{-\frac{3}{4} a^{3}\left(a^{4}\right)^{-3 / 4}}\\
&=\frac{-a-\frac{1}{3} a}{-\frac{3}{4}}\\
&=\frac{4}{3}\left(\frac{4}{3} a\right)\\
&=\frac{16}{9} a
\end{aligned}
$$
