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} $
Show Answer
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} $
