JEE Main 2021MathematicsLimitsEasyMCQ

JEE Main 2021 — Limits Question with Solution

JEE Main 2021 (27 Jul Shift 2)

Question

The value of limx→0x1-sinx8-1+sinx8 is equal to :

Choose an option

Show full solutionCorrect option: C
Correct answer
C-4

Step-by-step explanation

limx→0x1-sinx8-1+sinx8

=limx→0x1-sinx8-1+sinx8×1-sinx8+1+sinx81-sinx8+1+sinx8×

1-sinx4+1+sinx41-sinx4+1+sinx4×

1-sinx+1+sinx1-sinx+1+sinx

=limx→0x1-sinx-(1+sinx)×

1-sinx8+1+sinx81-sinx4+1+sinx41-sinx+1+sinx

=limx→0x(-2sinx)1-sinx8+1+sinx8×

1-sinx4+1+sinx41-sinx+1+sinx

=limx→0-12222 ∵ limx→0sinxx=1=-4

Practice this on the real CBT interface

Solve this JEE Main question (and the rest of the Limits chapter) on PrepSharp's TCS iON-style CBT player — with timer, bookmarks and session analytics.

Solve interactively →

About this question

This is a previous-year question from JEE Main 2021, 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.