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Let $f\colon(0,\infty)\to \mathbb R$ be a continuous function such that $f(x)\le f(nx) , \forall n \in \mathbb N , \forall x >0$ , then is it true that $\lim_{x\to \infty} f(x)$ exists (may be infinite) ?

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    $\begingroup$ Just a word of warning for you. If you continue deleting your questions before someone has taken the time and effort to help you, then the time will come when everyone refuses to answer you... I personally don't care if you ask homework-related questions, but I find that behavior utterly disrespectful. Just saying... $\endgroup$
    – user378947
    Dec 18, 2016 at 14:35

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Let $\ell=\sup f$. First assume that $\ell\in\mathbb{R}$. Let's show that $\lim\limits_{+\infty}f=\ell$. Let $\varepsilon>0$. By definition of $\ell$, there exists $x_0\in(0,+\infty)$ such that $f(x_0)>\ell-\varepsilon/2$. Since $f$ is continuous, there exists $\delta>0$ (with, of course, $\delta<x_0$ too) such that $$\forall x\in(x_0-\delta,x_0+\delta),\ f(x_0)-\varepsilon/2<f(x)<f(x_0)+\varepsilon/2,$$ and in particular, $$\forall x\in(x_0-\delta,x_0+\delta),\ \ell-\varepsilon<f(x)\leq\ell<\ell+\varepsilon.$$ Since $\delta>0$ there exists $n_0\in\mathbb{N}$ such that $x_0<n_0\delta$. Let $M=n_0(x_0+\delta)$ and let $x\in(M,+\infty)$. Let $k=\lfloor x/x_0\rfloor$ ($k$ is the integer part of $x/x_0$). By definition, $$k\leq\frac x{x_0}<k+1$$ but since $x>n_0(x_0+\delta)$ we conclude that $x/x_0>n_0+n_0\delta/x_0>n_0+1$, and hence $k>n_0$. Finally, we conclude that $$-k\delta<0\leq x-k x_0<x_0<n_0\delta<k\delta,$$ i.e., $$x\in(k x_0-k\delta,k x_0+k\delta).$$ Hence $f(x)=f(k x/k)\geq f(x/k)>\ell-\varepsilon$ since $x/k\in(x_0-\delta,x_0+\delta)$, and hence $$\ell-\varepsilon<f(x)\leq\ell<\ell+\varepsilon.$$

We hence showed that: $$\forall\varepsilon>0,\ \exists M>0,\ \forall x\in(M,+\infty),\ \ell-\varepsilon<f(x)<\ell+\varepsilon,$$ that is, $\lim\limits_{x\to+\infty}f(x)=\ell$.

In the case where $\ell\not\in\mathbb{R}$, i.e., $\ell=+\infty$, either repeat the previous proof with some straightforward adaptations, or consider the function $g=\arctan\circ f$. Then, since $\arctan$ is increasing, $g$ also satisfies the given relation; moreover, $\ell=\sup g=\pi/2\in\mathbb{R}$ so we can apply the previous proof to obtain $\lim\limits_{x\to+\infty}g(x)=\pi/2$; since the values of $g$ lie in $(-\pi/2,\pi/2)$ we conclude that $\lim\limits_{x\to+\infty}f(x)=+\infty$.

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