JEE Main 2019MathematicsLimitsEasyMCQ

JEE Main 2019 — Limits Question with Solution

JEE Main 2019 (12 Apr Shift 2)

Question

limx→0x+2sinxx2+2 sin x+1 - sin2x-x+1 is

Choose an option

Show full solutionCorrect option: C
Correct answer
C2

Step-by-step explanation

limx→0x+2sinxx2+2sinx+1−sinx−x+1   00form

By using L'Hospital Rule,

=limx→01+2cosx12x+2cosx2x2+2sinx+1−1sin2x−12sin2x−x+1

=1+21+12=332=2.

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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.