Limit of function $\lim_{x\to+\infty}\frac{f(x)}{\ln x}=0.$ I'm stuck in this question: If $f\colon \Bbb R^+\to\Bbb R^+$ is an increasing function and $a>1$ such that $\lim_{x\to+\infty}f(ax)-f(x)=0$, How to prove
$$\lim_{x\to+\infty}\frac{f(x)}{\ln x}=0.$$
Thanks.
 A: Let $\epsilon>0$, there's $A>0$ such that $|f(ax)-f(x)|\leq \epsilon$ if $x>A$.
Pick $x>A$. We have $\frac{x}{a^k}\geq A\iff k\leq \log_a \frac{x}{A}$, so let $n=\lfloor \log_a \frac{x}{A}\rfloor$ then
$$\left|f\left(\frac{x}{a^{k-1}}\right)-f\left(\frac{x}{a^{k}}\right)\right|\leq \epsilon,\, \forall k=1,\ldots,n\tag{1}$$
We sum the inequalities in $(1)$, since $f$ is increasing we have:
$$|f(x)|\leq \left|f(x)-f\left(\frac{x}{a^{n}}\right)\right|+\left|f\left(\frac{x}{a^{n}}\right)\right|\leq n\epsilon+|f(aA)|$$
Hence if we divide by $n=\lfloor \log_a \frac{x}{A}\rfloor$ we can make $\left|\frac{f(x)}{n}\right|$ as small as we want ($\leq 2\epsilon$) since $\lim_{x\to\infty}\frac{f(aA)}{n}=0$. Moreover we have $\lim_{x\to\infty}\frac{\log x}{n}=\log a$, hence finally for $B>0$ sufficently large we have
$$\left|\frac{f(x)}{\log x}\right|=\left|\frac{f(x)}{n}\right|\left|\frac{n}{\log x}\right|\leq\frac{2\epsilon(1+\epsilon)}{\log(a)}\leq \frac{3\epsilon}{\log(a)}$$
A: Let $x>N$  such that for all $x>N$
$f(ax)-f(x)\leq\epsilon$.
Then $f(a^n N)-f(N)\leq n\epsilon$ for $n \in \mathbb {N}$
If $t<a^nN$ and $t\geq a^{n-1}N$ with $n \in \mathbb{N}$ , and both $f$ and $\log$ are increasing,$$\dfrac{f(t)}{\log t}\leq \dfrac{f(a^n N)}{\log t}\leq\dfrac{f(a^nN)}{\log a^{n-1}N}\leq\dfrac{f(N)+n\epsilon}{(n-1)\log a+\log N}$$, which tends to 0(with $\epsilon$).
