JEE Main 2015MathematicsLimitsEasyMCQ

JEE Main 2015 — Limits Question with Solution

JEE Main 2015 (10 Apr Online)

Question

limx→0⁡ex2-cos⁡xsin2⁡ x is equal to

Choose an option

Show full solutionCorrect option: B
Correct answer
B32

Step-by-step explanation

limx→0ex2-cosxsin2x=limx→0(ex2−1)+(1−cosx)x2·sin2xx2

=limx→0ex2−1x2+1−cosxx2

=1+12=32

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About this question

This is a previous-year question from JEE Main 2015, covering the Limits chapter of Mathematics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.