Let [x] denote the greatest integer less than or equal to X . Then : _ x arrow 0 (π ^ 2 x)+(|x|- (x[x]))^ 2 x^ 2

Correct answer: A. does not exist RHL is, _ x arrow 0^ + (π ^ 2 x)+(x-0)^ 2 x^ 2 = _ x arrow 0^ + ( (π ^ 2 x. x^ 2 +1)=1+π And LHL is, _ x arrow 0^ - (π ^ 2 x)+(-x+ x)^ 2 x^ 2 = _ x arrow 0^ - (π ^ 2 x)+x^ 2 + ^ 2 x-2 x x x^ 2 =π+1+1-2=π Since, LHL ≠ RHL Hence, limit does not exist.

JEE Main 201911 JanuaryMorning Shift

Exam: JEE Main
Year: 2019
Date: 11 January
Shift: Morning Shift
Subject: Mathematics
Chapter: Limits

JEE Main 2019 · 11 January · Morning Shift · Mathematics · Limits

Let [x] denote the greatest integer less than or equal to X . Then...

Let

[\mathrm{x}]

denote the greatest integer less than or equal to

\mathrm{X}

. Then :

\lim _{x \rightarrow 0} \frac{\tan \left(\pi \sin ^{2} x\right)+(|\mathrm{x}|-\sin (x[x]))^{2}}{x^{2}}

Options

Adoes not exist
Bequals $\pi$
Cequals $\pi+1$
Dequals 0

Correct answer

A. does not exist

Step-by-step solution

RHL is,

\lim _{x \rightarrow 0^{+}} \frac{\tan \left(\pi \sin ^{2} x\right)+(x-0)^{2}}{x^{2}}

=\lim _{x \rightarrow 0^{+}}\left(\frac{\tan \left(\pi \sin ^{2} x\right.}{x^{2}}+1\right)=1+\pi

And LHL is,

\lim _{x \rightarrow 0^{-}} \frac{\tan \left(\pi \sin ^{2} x\right)+(-x+\sin x)^{2}}{x^{2}}

=\lim _{x \rightarrow 0^{-}} \frac{\tan \left(\pi \sin ^{2} x\right)+x^{2}+\sin ^{2} x-2 x \sin x}{x^{2}}

=\pi+1+1-2=\pi

Since, LHL

\neq

RHL Hence, limit does not exist.

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