Single Variable Calculus, Chapter 7, 7.8, Section 7.8, Problem 80

Find $\displaystyle \lim_{x \to a} \frac{\sqrt{2a^3 x - x^4} - a^3 \sqrt{a^2 x}}{a - \sqrt[4]{ax^3}}$
By applying L'Hospital's Rule...


$
\begin{equation}
\begin{aligned}
\lim_{x \to a} \frac{\sqrt{2a^3 x - x^4} - a^3 \sqrt{a^2 x}}{a - \sqrt[4]{ax^3}} &= \lim_{x \to a} \frac{\frac{1}{2}\left(2a^3x - x^4\right)^{-\frac{1}{2}} \left( 2a^3 - 4x^3 \right) - a \left(\frac{1}{3}\right)(a^2 x)^{\frac{-2}{3}}(a^2) }{0 - \frac{1}{4}(ax^3)^{\frac{3}{4}} (3ax^2)}\\
\\
&= \lim_{x \to a} \frac{\frac{2a^3 - 4x^3}{2\sqrt{2a^3 x - x^4}} - \frac{a^3}{3(a^2x)^{\frac{2}{3}}} }{-\frac{3ax^2}{4(ax^3)^{\frac{3}{4}}}}\\
\\
&= \lim_{x \to a} \frac{\frac{2(a^3-2x^3)}{2\sqrt{2a^3x-x^4}} - \frac{a^3}{3\left(a^{\frac{4}{3}} x^{\frac{2}{3}}\right)}}{-\frac{3}{4}\left[ a^{\left(1-\frac{3}{4}\right)} \cdot x^{\left( 2 - \frac{9}{4} \right)}\right]}\\
\\
&= \lim_{x \to a} \frac{\frac{a^3 - 2x^3}{\sqrt{2a^3x-x^4}} - \frac{a^{\left(3 - \frac{4}{3}\right)}}{3(x^{\frac{2}{3}})}}{-\frac{3}{4}\left[ a^{\frac{1}{4}} x^{\frac{-1}{4}}\right]}\\
\\
&= \lim_{x \to a} \frac{\frac{\left( 3x^{\frac{2}{3}} \right)\left( a^3 - 2x^3 \right) - a^{\frac{5}{3}} \left( \sqrt{2a^3x - x^4} \right) }{3 \left( x^{\frac{2}{3}} \right) \sqrt{2a^3x - x^4}} }{-\frac{3}{4}\left(\frac{a}{x}\right)^{\frac{1}{4}}}
\end{aligned}
\end{equation}
$


If we substitute the value of the limit,


$
\begin{equation}
\begin{aligned}
&= \frac{\frac{\left( 3x^{\frac{2}{3}} \right)\left( a^3 - 2x^3 \right) - a^{\frac{5}{3}} \left( \sqrt{2a^3x - x^4} \right) }{3 \left( x^{\frac{2}{3}} \right) \sqrt{2a^3x - x^4}} }{-\frac{3}{4}\left(\frac{a}{x}\right)^{\frac{1}{4}}}\\
\\
&= \frac{\frac{\left(3a^{\frac{2}{3}}\right) (-a^3) - a^{\frac{5}{3}} \sqrt{a^4} }{3\left( a^{\frac{2}{3}} \right) \sqrt{a^4}}}{- \frac{3}{4}(1)}\\
\\
&= \frac{\frac{-3a^{\frac{11}{3}} - a^{\frac{11}{3}} }{3a^{\frac{8}{3}}}}{-\frac{3}{4}}\\
\\
&= \frac{-4a^{\frac{11}{3}}}{3a^{\frac{8}{3}}} \left( -\frac{4}{3} \right)\\
\\
&= \frac{16}{9} a^{\frac{11}{8}- \frac{8}{3}}\\
\\
&= \frac{16}{9}a
\end{aligned}
\end{equation}
$