JEE Main 2018 (Online) 16th AprilQuestion 1
$\lim_{x\to0}\frac{(27+x)^{\frac13}-3}{9-(27+x)^{\frac23}}$ equals
Limits & Continuity
Solution
Put $y=(27+x)^{1/3}\to3$: $\frac{y-3}{9-y^2}=-\frac1{3+y}\to-\frac16$.Limits & Continuity - JEE Mathematics question (JEE Main 2018 (Online) 16th April)
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