Answer
$$
\lim _{x \rightarrow a} \frac{\sqrt{2 a^{3} x-x^{4}}-a \sqrt[3]{a a x}}{a-\sqrt[4]{a x^{3}}} =\frac{16}{9} a
$$
Work Step by Step
$$
\lim _{x \rightarrow a} \frac{\sqrt{2 a^{3} x-x^{4}}-a \sqrt[3]{a a x}}{a-\sqrt[4]{a x^{3}}}
$$
Since
$$
\lim _{x \rightarrow a} ( \sqrt{2 a^{3} x-x^{4}}-a \sqrt[3]{a a x})=0
$$
and
$$
\lim _{x \rightarrow a} ( a-\sqrt[4]{a x^{3}})=0
$$
the limit is an indeterminate form of type $\frac{0}{0}$ so we can apply l’Hospital’s Rule:
$$
\begin{aligned}
\lim _{x \rightarrow a} & \frac{\sqrt{2 a^{3} x-x^{4}}-a \sqrt[3]{a a x}}{a-\sqrt[4]{a x^{3}}} \stackrel{\mathrm{H}}{=} \\
& \lim _{x \rightarrow a} \frac{\frac{1}{2}\left(2 a^{3} x-x^{4}\right)^{-1 / 2}\left(2 a^{3}-4 x^{3}\right)-a\left(\frac{1}{3}\right)(a a x)^{-2 / 3} a^{2}}{-\frac{1}{4}\left(a x^{3}\right)^{-3 / 4}\left(3 a x^{2}\right)} \\
& =\frac{\frac{1}{2}\left(2 a^{3} a-a^{4}\right)^{-1 / 2}\left(2 a^{3}-4 a^{3}\right)-\frac{1}{3} a^{3}\left(a^{2} a\right)^{-2 / 3}}{-\frac{1}{4}\left(a a^{3}\right)^{-3 / 4}\left(3 a a^{2}\right)} \\
& =\frac{\left(a^{4}\right)^{-1 / 2}\left(-a^{3}\right)-\frac{1}{3} a^{3}\left(a^{3}\right)^{-2 / 3}}{-\frac{3}{4} a^{3}\left(a^{4}\right)^{-3 / 4}} \\
&=\frac{-a-\frac{1}{3} a}{-\frac{3}{4}} \\
&=\frac{4}{3}\left(\frac{4}{3} a\right) \\
&=\frac{16}{9} a.
\end{aligned}
$$
