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

Answer 1

The correct option is C $1$

Given: f is a +ve increasing sunction

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

To find:${lim}_{x\to \infty }\frac{f\left(2x\right)}{f\left(x\right)}$

Sol: $f\left(x\right)$ is a positive increasing function

$⟹f^{{}^{\prime }}\left(x\right)>0forallx$

$f\left(x\right)\le f\left(2x\right)\le f\left(3x\right)[\therefore increasingfunction]$

$⟹\frac{f\left(x\right)}{f\left(x\right)}\le \frac{f\left(2x\right)}{f\left(x\right)}\le \frac{f\left(3x\right)}{f\left(x\right)}[\therefore f\left(x\right)>0]$

${lim}_{x\to \infty }1\le {lim}_{x\to \infty }\frac{f\left(2x\right)}{f\left(x\right)}\le {lim}_{x\to \infty }\frac{f\left(3x\right)}{f\left(x\right)}\left[sandwichtheorem\right]$

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

Hence, correct answer is 1