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

Answer 1

Answer: (C)

$-\frac{1}{6}$

The given limit is in the indeterminate form $\frac{0}{0}$

Using the L'Hospital Rule we get:

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

Now, substituting the value of $x$ we get:

$=\frac{\frac{1}{3}\times (27{)}^{\frac{-2}{3}}}{\frac{-2}{3}\times {27}^{-\frac{1}{3}}}$

$=-\frac{1}{6}$