Question

The value of $\underset{x\to 1}{\lim }$$\frac{x^{1/4}-x^{1/5}}{x^3-1}$ is

• A. $\frac{1}{20}$
• B. $\frac{1}{40}$
• C. $\frac{1}{60}$
• D. $\frac{3}{20}$

Answer 1

Answer: (C)

$\underset{x\to 1}{\lim }$$\frac{x^{1/4}-x^{1/5}}{x^3-1}$
$=\underset{x\to 1}{\lim }$$\frac{\left(x^{1/4}-1\right)-\left(x^{1/5}-1\right)}{x^3-1}$
$=\underset{x\to 1}{\lim }$$\left(\frac{\frac{x^{1/4}-1^{1/4}}{x-1}}{\frac{x^3-1^3}{x-1}}-\frac{\frac{x^{1/5}-x^{1/5}}{x-1}}{\frac{x^3-1^3}{x-1}}\right)$
$=\frac{1}{12}-\frac{1}{15}=\frac{1}{60}$