Question

Let $\mathrm{f}:\mathrm{R}\to \mathrm{R}$ be a positive increasing function with $\underset{\mathrm{x}\to \infty }{\lim }\frac{\mathrm{f}\left(3\mathrm{x}\right)}{\mathrm{f}\left(\mathrm{x}\right)}1.$ Then, $\underset{\mathrm{x}\to \infty }{\lim }\frac{\mathrm{f}\left(2\mathrm{x}\right)}{\mathrm{f}\left(\mathrm{x}\right)}=1$ is equal to

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

Answer 1

The correct option is A

$1$

Explanation for the correct answer:

Since, $\mathrm{f}\left(\mathrm{x}\right)$ is increasing positive $x$, we have

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

Also, $\mathrm{f}\left(\mathrm{x}\right)>0$

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

Considering the value of the limit as $\mathrm{x}\to \infty ,$ by the Sandwich theorem

Let $f,g,\mathrm{and}h$ be real functions such that $f\left(x\right)\le g\left(x\right)\le h\left(x\right)$ for all $x$ in the common domain of definition.

If, $\underset{x\to a}{\lim }f\left(x\right)=l=\underset{x\to a}{\lim }h\left(x\right)\mathbf{\text{, then}}\underset{x\to a}{\lim }g\left(x\right)=l$

Now we have,

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

Hence, the correct answer is an option (A).