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Q. $\displaystyle\lim_{x \to 0} \frac{(27 + x)^{\frac{1}{3}} - 3}{9- (27 + x)^{\frac{2}{3}}}$ equals :

JEE MainJEE Main 2018Limits and Derivatives

Solution:

$\displaystyle\lim _{x \rightarrow 0} \frac{(27+x)^{\frac{1}{3}}-3}{9-(27+x)^{\frac{2}{3}}}$
$\displaystyle\lim _{x \rightarrow 0} \frac{3\left[\left(1+\frac{x}{27}\right)^{1 / 3}-1\right]}{9\left[1-\left(1+\frac{x}{27}\right)^{23}\right]} $
$\displaystyle\lim _{x \rightarrow 0} \frac{3\left[1+\frac{1}{3}\left(\frac{x}{27}\right)-1\right]}{9\left[1-1-\frac{2}{3}\left(\frac{x}{27}\right)\right]} $
$\displaystyle\lim _{x \rightarrow 0} \frac{1}{3}\left[\frac{\frac{x}{81}}{\frac{-2}{3} \cdot \frac{x}{27}}\right]=\frac{-1}{6}$