Question

The value of $\underset{x\to 64}{lim}\frac{x^{1/2}-8}{x^{1/3}-4}$ is

• A. $1$
• B. $2$
• C. $3$
• D. None of the above

Answer 1

Answer: (C)

$3$

Clearly given limit is of form $\frac{0}{0}$. so using L-hospital's rule,
$\underset{x\to 64}{lim}\frac{x^{1/2}-8}{x^{1/3}-4}=\underset{x\to 64}{lim}\frac{\frac{1}{2}{x}^{-1/2}}{\frac{1}{3}{x}^{-2/3}}$
$=\underset{x\to 64}{lim}\frac{3}{2}{x}^{1/6}=3$
Hence, option 'C' is correct.