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$f: (0,\infty) \rightarrow \mathbb{R}$ is continuous and $f(x) \le f(nx) \ \ \forall{{x>0},{n\in\mathbb{N}}} $

Prove that f has limit (may be infinity)

Don't know what to do maybe something with Darboux property.

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$f$ has limit when $x$ tends to... –  Tomás Jan 14 '13 at 21:14
    
I haven't specified in the exercise, assume that $x \rightarrow \infty$ –  aiki93 Jan 14 '13 at 21:40

1 Answer 1

$\textbf{Lemma}$: Let $0<a<b$. Then there exist $N>0$ such that $(N,\infty)\subseteq\cup_{n=1}^\infty (na,nb)$.

$\textbf{Proof}:$ Let $k_0$ be big enough such that $\frac{k+1}{k}<\frac{b}{a}$ for all $k\geq k_0$. Then $(ka,kb)$ and $((k+1)a,(k+1)b)$ have no trivial intersection for all $k\geq k_0$. Choose $N=k_0$.

Now suppose that there exist two sequences $\{x_n\}$ and $\{y_n\}$ such that $\lim x_n=\lim y_n=\infty$ and $\lim f(x_n)=x<y=\lim f(y_n)$ for some $x,y\in [-\infty,+\infty]$. Choose $x<w<y$ and let $n_0$ be such that $f(x_n)<w$ and $f(y_n)>w$ for all $n\geq n_0$. Let $a=y_{n_0}$ and $b=y_{n_0}+\delta$ for some $\delta>0$ such that $f(z)>w$ for all $z\in (y_{n_0},y_{n_0}+\delta)$. Using the lemma above we conclude that there exists $N$ such that $(N,\infty)\in \cup_{n=1}^\infty (na,nb)$ and this with the fact that $f(z)\leq f(nz)$ for all $n$ and for all $z$ imply that $f(z)>w$ for all $w>N$ which contradicts the fact that $f(x_n)<w$ for all $n>n_0$.

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