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

Answer: (D)

$1$

$x\le 2x\le 3x$

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

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

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

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

So, by Sandwich/ Squeeze Play Theorem of limit, we can conclude that,

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