Question

Let $f : R \to R$ be a positive increasing function with $\displaystyle \lim_{x \to \infty} \frac{f\left(3x\right)}{f\left(x\right)} =1$
Then $\displaystyle \lim_{x \to\infty} \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)

f(x) is a positive increasing function.
$\therefore \, 0 < f(x) < f(2x) < f(3x)$
$\Rightarrow \, 0 < 1 < \frac{f(2x)}{f(x)} < \frac{f(3x)}{f(x)}$
$\Rightarrow \displaystyle \lim _{x \to \infty } 1\le \displaystyle \lim_{x \to \infty} \frac{f\left(2x\right)}{f\left(x\right)} \le \displaystyle \lim_{x \to\infty} \frac{f\left(3x\right)}{f\left(x\right)}$
By Sandwich Theorem.
$\Rightarrow \displaystyle \lim _{x \to \infty } \frac{f\left(2x\right)}{f\left(x\right)}=1$