Question

$\underset{x\to 0}{\lim }\frac{(27+x{)}^{\frac{1}{3}}-3}{9-(27+x{)}^{\frac{2}{3}}}$ equals :

• A. $\frac{1}{3}$
• B. $-\frac{1}{3}$
• C. $-\frac{1}{6}$
• D. $\frac{1}{6}$

Answer 1

Answer: (C)

$\underset{x\to 0}{\lim }\frac{(27+x{)}^{\frac{1}{3}}-3}{9-(27+x{)}^{\frac{2}{3}}}$
$\underset{x\to 0}{\lim }\frac{3\left[{\left(1+\frac{x}{27}\right)}^{1/3}-1\right]}{9\left[1-{\left(1+\frac{x}{27}\right)}^{23}\right]}$
$\underset{x\to 0}{\lim }\frac{3\left[1+\frac{1}{3}\left(\frac{x}{27}\right)-1\right]}{9\left[1-1-\frac{2}{3}\left(\frac{x}{27}\right)\right]}$
$\underset{x\to 0}{\lim }\frac{1}{3}\left[\frac{\frac{x}{81}}{\frac{-2}{3}\cdot \frac{x}{27}}\right]=\frac{-1}{6}$