JEE Main 2015 (Online) 10th AprilQuestion 1
$\lim_{x\to0}\frac{e^{x^2}-\cos x}{\sin^2x}$ is equal to
Limits & Continuity
Solution
$e^{x^2}-\cos x\approx x^2+\frac{x^2}2=\frac32x^2$ and $\sin^2x\approx x^2$, so the limit is $\frac32$.Limits & Continuity - JEE Mathematics question (JEE Main 2015 (Online) 10th April)
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