JEE Main 2019MathematicsLimitsEasyMCQ

JEE Main 2019 — Limits Question with Solution

JEE Main 2019 (08 Apr Shift 1)

Question

limx→0 sin2x2-1+cosx equals

Choose an option

Show full solutionCorrect option: A
Correct answer
A42

Step-by-step explanation

limx→0 sin2x2-1+cosx

On Rationalisation, we get,

     =limx→0 sin2x2-1+cosx×2+1+cosx2+1+cosx

=limx→0 2sinx2cosx221-cosx×2+1+cosx

=limx→0 4sin2x2cos2x22sin2x2×2+1+cosx

=limx→0 2cos2x22+1+cosx=2×22

=42.

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About this question

This is a previous-year question from JEE Main 2019, 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.