Mathematics · 2018
JEE Main · 16 April 2018, Shift 1 · Q70
lim_x → 0 ((27+x)^(1/3)-3)/(9-(27+x)^(2/3)) equals:
$\displaystyle \lim _{x \rightarrow 0} \frac{(27+x)^{\frac{1}{3}}-3}{9-(27+x)^{\frac{2}{3}}}$ equals :
Official answer
(3)
$\displaystyle -\frac{1}{6}$
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JEE Main 2018 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.