Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Let be a positive increasing function with <img alt="\mat

Let \mathrm{f: R \rightarrow R} be a positive increasing function with \mathrm{ \lim _{x \rightarrow \infty} \frac{f(3 x)}{f(x)}=1. Then, \lim _{x \rightarrow \infty} \frac{f(2 x)}{f(x)}=}
 

Option: 1

\frac{3}{2}


Option: 2

3


Option: 3

1


Option: 4

\frac{2}{3}


Answers (1)

best_answer

It is given that \mathrm{f(x)} is a positive increasing function.

\mathrm{ \therefore 0<f(x)<f(2 x)<f(3 x) \\ }

\mathrm{ \Rightarrow 0<1<\frac{f(2 x)}{f(x)}<\frac{f(3 x)}{f(x)} \Rightarrow 1<\frac{f(3 x)}{f(x)} }

Clearly, \mathrm{ \lim _{x \rightarrow \infty} 1=1\: and \: \lim _{x \rightarrow \infty} \frac{f(3 x)}{f(x)}=1 }.

Therefore, by Sandwich theorem, we have \mathrm{ \lim _{x \rightarrow \infty} \frac{f(2 x)}{f(x)}=1. }

Hence option 3 is correct.

Posted by

avinash.dongre

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2026: IIT Roorkee’s ECE emerges as top choice after CSE; closing rank rises
anam-khan

JEE Main 2026 application correction begins on jeemain.nta.nic.in; edit details by December 2
anam-khan

JEE Mains 2026 LIVE: NTA to conduct session 1 from January 21 to 30; exam pattern, syllabus

Latest Question