Question

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

• A. $-1/6$
• B. $1/6$
• C. $1/3$
• D. $-1/3$

Answer 1

The correct option is A $-1/6$

$\underset{x\to 0}{lt}\frac{{(27+x)}^{1/3}-3}{9-{(27+x)}^{2/3}}$

$=\underset{x\to 0}{lt}\frac{\frac{1}{3}{(27+x)}^{-2/3}}{\frac{-2}{3}{(27+x)}^{-1/3}}$

$=\underset{x\to 0}{lt}\frac{-1}{2}\frac{{(27+x)}^{1/3}}{{(27+x)}^{2/3}}$

$=\frac{-1}{2}\left(\frac{3}{9}\right),\frac{-1}{6}$ [A]