JEE Main 2020MathematicsLimitsMediumMCQ

JEE Main 2020 — Limits Question with Solution

JEE Main 2020 (03 Sep Shift 2)

Question

limx→aa+2x13-3x133a+x13-4x13 a≠0 is equal to:

Choose an option

Show full solutionCorrect option: D
Correct answer
D232913

Step-by-step explanation

At x=a, limit is of 00 form.

So, applying L'Hospital rule, we get

limx→aa+2x13-3x133a+x13-4x13=limx→a13a+2x-23.2-13.3x-23.3133a+x-23.-13.4x-23.4

=133a-23.2-3134a-23.1-4=3-234-23.13

=243913.13=232913

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 2020, 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.