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Limits Continuity and Differentiability question

2010 · Shift 0 · Q28

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JEE Main
Mathematics
Limits Continuity and Differentiability
2010 · Shift 0 · Q28

Limits Continuity and Differentiability question

2010 · Shift 0 · Q28

JEE MainMathematicsLimits Continuity and DifferentiabilityMCQ+4 / −1

Let

f:R \to R

be a positive increasing function with

\mathop {\lim }\limits_{x \to \infty } {{f(3x)} \over {f(x)}} = 1

. Then

\mathop {\lim }\limits_{x \to \infty } {{f(2x)} \over {f(x)}} =

A
${2 \over 3}$
B
${3 \over 2}$
C
3
D
1

View written solutionFree

Correct answer: D

Given condition

We are given that $f:\mathbb{R}\to\mathbb{R}$ is positive and increasing, and

\lim_{x\to\infty}\frac{f(3x)}{f(x)}=1.

We need to find

\lim_{x\to\infty}\frac{f(2x)}{f(x)}.

Use monotonicity

Since $f$ is increasing and $x<2x<3x$ for all large positive $x$ , we have

f(x)\le f(2x)\le f(3x).

Because $f(x)>0$ , divide throughout by $f(x)$ :

1\le \frac{f(2x)}{f(x)}\le \frac{f(3x)}{f(x)}.

Take limits

We know

\lim_{x\to\infty}\frac{f(3x)}{f(x)}=1.

So from

1\le \frac{f(2x)}{f(x)}\le \frac{f(3x)}{f(x)},

by the Squeeze Theorem,

\lim_{x\to\infty}\frac{f(2x)}{f(x)}=1.

Check options

A: $\frac23$ — incorrect
B: $\frac32$ — incorrect
C: $3$ — incorrect
D: $1$ — correct

Thus the correct answer is

\boxed{1}.

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