Question

$\underset{x\to 0}{lim}\frac{x\sqrt[3]{3z^2-(z-x{)}^2}}{(\sqrt[3]{38xz-4x^2}+\sqrt[3]{38xz}{)}^4};(z\ne 0)$ is equal to

• A. $\frac{z}{2^{11/3}}$
• B. $\frac{1}{2^{23/3}\cdot z}$
• C. $\frac{1}{2^{21/3}\cdot z}$
• D. $\frac{z}{2^{21/3}}$

Answer 1

The correct option is B $\frac{1}{2^{23/3}\cdot z}$

$\underset{x\to 0}{lim}\frac{x\sqrt[3]{3z^2-(z-x{)}^2}}{(\sqrt[3]{38xz-4x^2}+\sqrt[3]{38xz}{)}^4}$
This is $\frac{0}{0}$ form, solve by simplification
$\underset{x\to 0}{lim}\frac{x\sqrt[3]{3z^2-(z-x{)}^2}}{(\sqrt[3]{38xz-4x^2}+\sqrt[3]{38xz}{)}^4}=\underset{x\to 0}{lim}\frac{x^{4/3}\sqrt[3]{32z-x}}{x^{4/3}(\sqrt[3]{38z-4x}+\sqrt[3]{38z}{)}^4}=\frac{\sqrt[3]{32z}}{[2\sqrt[3]{38z}{]}^4}=\frac{1}{2^{23/3}\cdot z}$