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

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

Answer 1

The correct option is D $1$

$\sin cex<2x<3x$

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

$1\le \frac{f\left(2x\right)}{f\left(x\right)}\le \frac{f\left(3x\right)}{f\left(x\right)}$

$\Rightarrow 1\le \underset{x\to \infty }{lim}\frac{f\left(x\right)}{f\left(x\right)}\le \underset{x\to \infty }{lim}\frac{f\left(3x\right)}{f\left(x\right)}=1$

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