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## Asymptotic characterization in terms of sequence limits

Suppose ${f:(0,\infty) \rightarrow \Bbb{R}}$ is a continuous function such that for every ${x>0}$ we have

$\displaystyle \lim_{n \rightarrow \infty} f(nx)=0.$

Prove that ${\lim\limits_{x \rightarrow \infty} f(x)=0}$.

Solution: In problems like this Baire’s theorem almost suggests itself. For some ${\varepsilon>0}$ define the following sets

$\displaystyle A_n = \bigcap_{m \geq n}^\infty \{x : |f(mx)| \leq \varepsilon \}$

and note that the hypothesis implies that for every ${x >0}$ there exists ${n_x \in \Bbb{N}}$ such that ${x \in A_{n_x}}$. This proves that ${\bigcup_{n=1}^\infty A_n =(0,\infty)}$. Moreover, from the definition we have ${A_n \subset A_p}$ for ${n\leq p}$. The continuity of ${f}$ assures us that the sets ${A_n}$ are closed.

The Baire category theorem implies that at least one ${A_n}$ has non-void interior and that means that there exists a non-trivial interval ${I}$ such that

$\displaystyle \forall x \in I, \forall m \geq n \Rightarrow |f(mx)|\leq \varepsilon.$

The main idea now is to realize that ${\bigcup_{m \geq n} mI}$ contains every real number which is sufficiently large. To do this, suppose ${I=(a,b)}$ with ${a and note that the inequality ${nb<(n+1)a}$ can only hold for a finite number of values for ${n}$.

Thus we have proved that for ${\varepsilon>0}$ there exists ${K_\varepsilon}$ such that ${|f(x)|\leq \varepsilon}$ for every ${x \geq K_\varepsilon}$. This is exactly the definition of the fact that the limit of ${f}$ at ${\infty}$ is zero.