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

$$\begin{gathered} \mathrm{clearly} \mathrm{x}<2\mathrm{x}<3\mathrm{x} \mathrm{whenx}>0 \\ \Rightarrow \mathrm{f}\left(\mathrm{x}\right)<\mathrm{f}\left(2\mathrm{x}\right)<\mathrm{f}\left(3\mathrm{x}\right) \\ \therefore \mathrm{f} \mathrm{is} \mathrm{increas}ing \mathrm{function} \\ \Rightarrow 1<\frac{\mathrm{f}\left(2\mathrm{x}\right)}{\mathrm{f}\left(\mathrm{x}\right)}<\frac{\mathrm{f}\left(3\mathrm{x}\right)}{\mathrm{f}\left(\mathrm{x}\right)} \\ li{m}_{x\to \infty }\left(1\right)<li{m}_{x\to \infty }\left(\frac{\mathrm{f}\left(2\mathrm{x}\right)}{\mathrm{f}\left(\mathrm{x}\right)}\right)<li{m}_{x\to \infty }\left(\frac{\mathrm{f}\left(3\mathrm{x}\right)}{\mathrm{f}\left(\mathrm{x}\right)}\right) \\ \Rightarrow 1<li{m}_{x\to \infty }\left(\frac{\mathrm{f}\left(2\mathrm{x}\right)}{\mathrm{f}\left(\mathrm{x}\right)}\right)<1 \\ By sandwich theorem li{m}_{x\to \infty }\left(\frac{\mathrm{f}\left(2\mathrm{x}\right)}{\mathrm{f}\left(\mathrm{x}\right)}\right)=1 \\ \therefore li{m}_{x\to \infty }\underset{ }{\mathrm{ }}\frac{\mathrm{f}\left(2\mathrm{x}\right)}{\mathrm{f}\left(\mathrm{x}\right)}=1 \end{gathered}$$