Question

The value of $\underset{x\to 0}{lim}\frac{(16+5x{)}^{1/4}-2}{(32+3x{)}^{1/5}-2}$ is

• A. $\frac{4}{5}$
• B. $\frac{25}{6}$
• C. $\frac{3}{8}$
• D. none of these

Answer 1

The correct option is B $\frac{25}{6}$

Let $L=\underset{x\to 0}{lim}\frac{(16+5x{)}^{1/4}-2}{(32+3x{)}^{1/5}-2}$

Clearly given limit is of the form $\frac{0}{0}$, thus using L-hospital rule,

$L=\underset{x\to 0}{lim}\frac{\frac{1}{4}.5(16+5x{)}^{-3/4}}{\frac{1}{5}.3(32+3x{)}^{-4/5}}$

$=\frac{25\times 32}{12\times 16}$

$=\frac{25}{6}$

Hence, option 'B' is correct.