## Calculus: Early Transcendentals (2nd Edition)

The solution is $$\lim_{x\to a}\frac{\sqrt{2a^3x-x^4}-a\sqrt[3]{a^2x}}{a-\sqrt[4]{ax^3}}=\frac{16}{9}a.$$
We will apply L'Hopital's rule to solve this limit. "LR" will stand for "Apply L'Hopital's rule". $$\lim_{x\to a}\frac{\sqrt{2a^3x-x^4}-a\sqrt[3]{a^2x}}{a-\sqrt[4]{ax^3}}=\left[\frac{\sqrt{2a^4-a^4}-a\sqrt[3]{a^3}}{a-\sqrt[4]{a^4}}\right]=\left[\frac{a^2-a^2}{a-a}\right]=\left[\frac{0}{0}\right][\text{LR}]=\lim_{x\to a}\frac{(\sqrt{2a^3x-x^4}-a\sqrt[3]{a^2x})'}{(a-\sqrt[4]{ax^3})'}=\lim_{x\to a}\frac{\frac{1}{2\sqrt{2a^3x-x^4}}(2a^3x-x^4)'-a\frac{1}{3(a^2x)^\frac{2}{3}}(a^2x)'}{-\frac{1}{4(ax^3)^\frac{3}{4}}(ax^3)'}=\lim_{x\to a}\frac{\frac{2a^3-4x^3}{2\sqrt{2a^3x-x^4}}-\frac{a^3}{3(a^2x)^\frac{2}{3}}}{-\frac{3ax^2}{4(ax^3)^\frac{3}{4}}}=\lim_{x\to a}\frac{\frac{a^3-2x^3}{\sqrt{2a^3x-x^4}}-\frac{1}{3}\left(\frac{a^5}{x^2}\right)^\frac{1}{3}}{-\frac{3}{4}\left(\frac{a}{x}\right)^\frac{1}{4}}=\left[\frac{\frac{a^3-2a^3}{\sqrt{2a^4-a^4}}-\frac{1}{3}\left(\frac{a^5}{a^2}\right)^\frac{1}{3}}{-\frac{3}{4}\left(\frac{a}{a}\right)^\frac{1}{4}}\right]=\left[\frac{-\frac{4}{3}a}{-\frac{3}{4}}\right]=\frac{16}{9}a.$$