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

By using sandwich theorem
  Since f is increasing function 
$\begin{array}{l}=f\left(x\right)<f\left(2x\right)<f\left(3x\right)\\ =\frac{f\left(x\right)}{f\left(x\right)}<\frac{f\left(2x\right)}{f\left(x\right)}<\frac{f\left(3x\right)}{f\left(x\right)}\\ =\underset{x\to 0}{\lim }\frac{f\left(x\right)}{f\left(x\right)}<\underset{x\to 0}{\lim }\frac{f\left(2x\right)}{f\left(x\right)}<\underset{x\to 0}{\lim }\frac{f\left(3x\right)}{f\left(x\right)}\\ =1<\underset{x\to 0}{\lim }\frac{f\left(2x\right)}{f\left(x\right)}<1\\ fromsandwitchtherom\\ \underset{x\to 0}{\lim }\frac{f\left(2x\right)}{f\left(x\right)}=1\end{array}$