Question

Let $f:R\to R$ be a positive increasing function with $\underset{x\to \infty }{lim}\frac{f\left(3x\right)}{f\left(x\right)}=1$ . Then , $\underset{x\to \infty }{lim}\frac{f\left(2x\right)}{f\left(x\right)}=1$ is equal to

• A. $1$
• B. $\frac{2}{3}$
• C. $\frac{3}{2}$
• D. $3$

Answer 1

The correct option is A $1$

Since, $f\left(x\right)$ is increasing for positive $x$, we have

$f\left(x\right)\le f\left(2x\right)\le f\left(3x\right)$

Also,

$f\left(x\right)>0$

$\Rightarrow \frac{f\left(x\right)}{f\left(x\right)}\le \frac{f\left(2x\right)}{f\left(x\right)}\le \frac{f\left(3x\right)}{f\left(x\right)}$

Considering the value of the limit as $x\to \infty$, by sandwich theorem, we have

$\underset{x\to \infty }{lim}\frac{f\left(2x\right)}{f\left(x\right)}=1$

Hence, this is the required result.