Q53 / 90 · JEE Main 2010 · Mathematics
Let $f: R \rightarrow R$ be a positive increasing function with $\lim _{x \rightarrow \infty} \frac{f(3 x)}{f(x)}=1$. Then $\lim _{x \rightarrow \infty} \frac{f(2 x)}{f(x)}=$
- A$\frac{2}{3}$
- B$\frac{3}{2}$
- C3
- Correct answerD1
Solution
$f(x)$ is a positive increasing function
$$
\begin{aligned}
& \Rightarrow 0 < f(x) < f(2 x) < f(3 x) \\
& \Rightarrow 0 < 1 < \frac{f(2 x)}{f(x)} < \frac{f(3 x)}{f(x)} \\
& \Rightarrow \lim _{x \rightarrow \infty} 1 \leq \lim _{x \rightarrow \infty} \frac{f(2 x)}{f(x)} \leq \lim _{x \rightarrow \infty} \frac{f(3 x)}{f(x)}
\end{aligned}
$$
By sandwich theorem.
$$
\Rightarrow \lim _{x \rightarrow \infty} \frac{f(2 x)}{f(x)}=1
$$
Hence, the correct answer is option: 'D'.