# The limit of a function approaching infinity [duplicate]

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Suppose $f\in C(0,+\infty)$,and $\forall x \in (0,+\infty)$,we have \begin{align} \lim_{n\to \infty}f(nx)=0 \end{align} where $n$ is positive integer.How to show that $\lim_{x\to\infty}f(x)=0$?

I have tried proof by contradiction, if not, $\exists \epsilon>0$, such that $\forall M>0$,$\exists y>M$, s.t. $|f(y)|>\epsilon$. But I got stuck here because I have no idea how to use the continuity to get contradiction.

Thanks for your attention.

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## marked as duplicate by Davide Giraudo, Stefan Hansen, vonbrand, muzzlator, Dominic MichaelisApr 7 '13 at 15:25

this might help you math.stackexchange.com/questions/63870/… –  jim Jan 1 '13 at 6:45
@jim Thank you for your reply! But this time restriction that $\mathbb{R}^+\to\mathbb{R}^+$ is turned into $\mathbb{R}^+\to\mathbb{R}$. –  Golbez Jan 1 '13 at 6:54
The change in restriction doesn't matter,thanks jim! BTW, is there other proof without using Baire Category Theorem? –  Golbez Jan 1 '13 at 7:00
See this post. In particular Florian's answer gives a link to a proof that doesn't appeal to Baire. –  David Mitra Jan 1 '13 at 12:39

By definition, we can rephrase $\lim_{n \to \infty} f(nx) = 0$ as "For all $x > 0$, for any $\epsilon > 0$, there exists $N \in \mathbb{N}$ such that for any $n \ge N$, $|f(nx)| < \epsilon$."
Consider the special case $x = 1$. Then we have: for any $\epsilon > 0$, there exists $N \in \mathbb{N}$ such that for any $n \ge N$, $|f(n)| < \epsilon$. By definition, this means that $\lim_{n \to \infty} f(n) = 0$. Rename your variable $n$ as $x$ and we have $\lim_{x \to \infty} f(x) = 0$.
I think the first step needs further discussion. fix $x$ there exists $N$ such that $|f(nx)|<\epsilon$ for all $n>N$ cannot be changed directly into $\forall x$, there exist $N$, such that $|f(nx)|<\epsilon$. Secondly, $\lim_{n\to\infty}f(n)=0$ doesn't mean $\lim_{x\to\infty}f(x)=0$ –  Golbez Jan 1 '13 at 7:27
I'm afraid I don't understand your disagreement with the first step. You claimed that for all $x > 0$ we have $\lim_{n \to \infty} f(nx) = 0$, yes? I have simply expanded the definition of the limit. As for your second point, it is certainly trivially true that $\lim_{n \to \infty} f(n) = 0 \Leftrightarrow \lim_{x \to \infty} f(x) = 0$. –  GMB Jan 1 '13 at 7:32
@LevDub I don't see how $\lim_{n \to \infty} f(n) = 0 \Leftrightarrow \lim_{x \to \infty} f(x) = 0$. Define $f(x) = e^{k-n}$ if $x=n\pi^k$ for some positive integers $n,k$, and $f(x)=0$ otherwise. Then $\lim_{n\to\infty}f(nx)=0$ for all $x$ and $\lim_{n\to\infty}f(n)=0$ in particular, but $f(x)$ does not converge to zero because $f(k\pi^k)=1$ for all $k$. The condition that $f$ is continuous is essential and I don't see it applied in your proof attempt. –  user1551 Jan 1 '13 at 11:45
The first argument is that although each $x$, there exists an $N(x)$ such that $|f(N(x)x)|<\epsilon$, there is a probability that $\sup N(x)=+\infty$.The general $N$ for all $x$ should be constructed, it's not obvious. –  Golbez Jan 1 '13 at 13:34
@user1551 : It's certainly true that $lim_{n \to \infty} f(x, n) = 0$ does NOT imply that $lim_{x \to \infty} f(x, n) = 0$, as your example shows. But that's not what I've claimed. I've said something much simpler: that $\lim_{x \to \infty} f(x, n) = \lim_{y \to \infty} f(y, n)$. Note that $x$ and $y$ fill the same parameter slot, so they can't be treated as fundamentally different variables like you've done in your example. Surely you agree with that equality? Golbez : I haven't claimed that there is one $N$ that applies to all $x$. Rather, I've said that to each $x$ there is an $N$. –  GMB Jan 1 '13 at 19:49