JEE Main On 16 April 2018 Question 18

Question: $ \underset{x\to 0}{\mathop{\lim }},\frac{{{(27+x)}^{\frac{1}{3}}}-3}{9-{{(27+x)}^{^{\frac{2}{3}}}}} $ equals. [JEE Main 16-4-2018]

Options:

A) $ -\frac{1}{3} $

B) $ \frac{1}{6} $

C) $ -\frac{1}{6} $

D) $ \frac{1}{3} $

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Answer:

Correct Answer: C

Solution:

  • 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} $