Let $f:\mathbb{R}\to \mathbb{R}$ be a positive increasing function with ${\lim }_{x\to \infty }\frac{f(3x)}{f(x)}=1$. Then ${\lim }_{x\to \infty }\frac{f(2x)}{f(x)}=$

Since $f(x)$ is an increasing function, for $x>0$:$f(x)\le f(2x)\le f(3x)$.Divide by $f(x)$ (which is positive):$1\le \frac{f(2x)}{f(x)}\le \frac{f(3x)}{f(x)}$.Taking limits as $x\to \infty$:${\lim }_{x\to \infty }1\le {\lim }_{x\to \infty }\frac{f(2x)}{f(x)}\le {\lim }_{x\to \infty }\frac{f(3x)}{f(x)}$.$1\le {\lim }_{x\to \infty }\frac{f(2x)}{f(x)}\le 1$.By Sandwich Theorem, the limit is $1$.