# Prove the limit at infinity is 0

Suppose $f$ is continuous. For all $x>0$, the limit of $f(nx)$ when $n$ goes to infinity is $0$.

Then please prove that the limit of $f(x)$ as $x$ goes to infinity is $0$. (I totally stuck at it) I think it suffices to show that f is uniformly continuous on $[0, \infty)$.

• So for $x =1$ you have $\lim_{n\to \infty} f(n) = 0$? Mar 21 '13 at 15:45

Let $\epsilon >0$. Let $B_n$ be the set of positive reals numbers $x$ such that $\mid f(mx)\mid\leq\epsilon$ for all $m\geq n$. The hypothesis says that $(0,\infty)=\cup B_n$. The Baire category theorem implies that there exists $n$ such that $B_n$ contains an interval, say $(a,b)\subset B_n$. Then $\mid f(x)\mid\leq \epsilon$ for all $x\in \cup_{n\geq m} (na,nb)$. It is an easy exercise to prove that $\cup_{n\geq m} (na,nb)$ contains $(N,\infty)$ for some large $N$. Then $\mid f(x)\mid\leq \epsilon$ for all $x\geq N$.
Since the real numbers are closed under multiplication, it suffices to show that there is some $n$ for which $x_1 = nx_2$, therefore $x_1$ can be represented as $nx_2$.