Question

The value of $\underset{x\to 1}{lim}\frac{\sqrt[3]{3x}-1}{\sqrt[4]{4x}-1}$

• A. $\frac{3}{4}$
• B. $\frac{4}{3}$
• C. $-\frac{3}{4}$
• D. $-\frac{4}{3}$

Answer 1

The correct option is B $\frac{4}{3}$

Now,

$\underset{x\to 1}{lim}\frac{\sqrt[3]{3x}-1}{\sqrt[4]{4x}-1}$ $\left(\frac{0}{0}\right)$ form .

Now applying L'Hospital's rule we get,

$=\underset{x\to 1}{lim}\frac{\frac{1}{3}{x}^{-\frac{2}{3}}}{\frac{1}{4}{x}^{-\frac{3}{4}}}$

$=\frac{4}{3}$.