Question

Let $f:R\to R$ be a positive increasing function with $\underset{x\to \infty }{lim}\frac{f\left(7x\right)}{f\left(x\right)}=1.$ Then which of the following(s) is(are) correct?

• A. $\underset{x\to \infty }{lim}\frac{f\left(2x\right)}{f\left(x\right)}=1$
• B. $\underset{x\to \infty }{lim}\frac{f\left(3x\right)}{f\left(x\right)}=3$
• C. $\underset{x\to \infty }{lim}\frac{f\left(4x\right)}{f\left(x\right)}=1$
• D. $\underset{x\to \infty }{lim}\frac{f\left(5x\right)}{f\left(x\right)}=5$

Answer 1

Answer: (A), (C)

$\underset{x\to \infty }{lim}\frac{f\left(4x\right)}{f\left(x\right)}=1$

As, $f\left(x\right)$ is an increasing function,
$\Rightarrow f\left(x\right)\le f\left(2x\right)\le f\left(3x\right)\le \cdots \le f\left(7x\right)$
$\Rightarrow \underset{x\to \infty }{lim}\frac{f\left(x\right)}{f\left(x\right)}\le \underset{x\to \infty }{lim}\frac{f\left(2x\right)}{f\left(x\right)}\le \underset{x\to \infty }{lim}\frac{f\left(3x\right)}{f\left(x\right)}\le \cdots \le \underset{x\to \infty }{lim}\frac{f\left(7x\right)}{f\left(x\right)}$
$\Rightarrow 1\le \underset{x\to \infty }{lim}\frac{f\left(2x\right)}{f\left(x\right)}\le \underset{x\to \infty }{lim}\frac{f\left(3x\right)}{f\left(x\right)}\le \cdots \le 1$

From sandwich theorem :
$\underset{x\to \infty }{lim}\frac{f\left(2x\right)}{f\left(x\right)}=\underset{x\to \infty }{lim}\frac{f\left(3x\right)}{f\left(x\right)}=\cdots =\underset{x\to \infty }{lim}\frac{f\left(6x\right)}{f\left(x\right)}=1$