A) \[-\frac{1}{3}\]
B) \[\frac{1}{6}\]
C) \[-\frac{1}{6}\]
D) \[\frac{1}{3}\]
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: \[=\frac{\frac{1}{3}\times {{(27)}^{\frac{-2}{3}}}}{\frac{-2}{3}\times {{27}^{-\frac{1}{3}}}}\] \[=-\frac{1}{6}\]
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