# $\lim\limits_{n\to\infty}f\left(\frac{x}{n}\right)=0$ for every $x > 0$. Prove $\lim\limits_{x \to 0}f(x)=0$

Function $$f: (0, \infty) \to \mathbb{R}$$ is continuous. For every positive $$x$$ we have $$\lim\limits_{n\to\infty}f\left(\frac{x}{n}\right)=0$$. Prove that $$\lim\limits_{x \to 0}f(x)=0$$. I have tried to deduce something from definition of continuity, but with no effect.

• What level course is this for? To me this is the kind of question that seems like a Baire Category Theorem thing. – Moya Jan 7 '19 at 22:41
• $n$ is an integer? – leonbloy Jan 7 '19 at 22:46
• But can you just not rewrite the limit such that $y=\frac{x}{n}$ and so you get what you need by letting $y$ going to 0, which does happen as $n$ goes to infinity and $x$ is any positive real and f is continuous of course so the limit exists? – Marco Bellocchi Jan 7 '19 at 22:46
• @Moya introductory level – user4201961 Jan 7 '19 at 22:48
• This I believe is just a bit more general than the following, you can proceed in a similar manner math.stackexchange.com/questions/3045776/… – Marco Bellocchi Jan 7 '19 at 22:59

## 2 Answers

The proof relies on the following lemma:

Let $$I \subset (0,\infty)$$ be a closed bounded interval. Let $$U \subset (0,\infty)$$ be an open subset accumulating at $$0$$. Then there exists some integer $$N \geq 2$$ and some closed interval $$J \subset U$$ such that $$N \cdot J \subset I$$.

Sketch of proof: take some $$x \in U$$ with $$x$$ lower than the length of $$I$$ and the minimum of $$I$$ . For some integer $$N \geq 2$$, $$Nx \in \overset{\circ}{I}$$, thus, there exists some compact interval $$J \subset U$$ such that $$N \cdot J \subset I$$.

Now, assume $$f$$ does not go to $$0$$ at $$0$$. Then, for some $$\epsilon > 0$$, $$U=\{|f| > \epsilon\}$$ is an open subset of $$(0,\infty)$$ accumulating at $$0$$.

By iterating the lemma, we can construct sequences of compact intervals $$(I_n)$$ and integers $$(N_n)$$ such that $$N_{n+1}I_{n+1} \subset I_{n-1}$$, $$N_n \geq 2$$.

Now, let $$A_n=N_1 \ldots N_n$$, then $$A_n \rightarrow \infty$$ and $$K_n=A_n \cdot I_n$$ is a non-increasing sequence of nonempty compact intervals in $$I_0$$. Thus, there exists $$x$$ that is in every $$K_n$$.

So, $$x/A_n \in I_n \subset U$$, for all $$n$$, thus $$|f(x/A_n)| > \epsilon$$ for all $$n$$, a contradiction.

• (edited) This looks correct, but I think you can remove some of the intermediate claims without affecting the argument. In particular, you don't need any information about $\sup I_n$. Also, when you say the sequence $K_n$ is non-decreasing, I think you mean that it is decreasing, or in more familiar language, descending. And in the final sentence, "all $n$" should be "infinitely many $n$". – Chris Culter Jan 7 '19 at 23:41
• Thanks for the remarks, I edited. However, in the last sentence, I definitely meant « for all $n$ » and not « for infinitely many $n$», however this does not affect the result. – Mindlack Jan 8 '19 at 0:02
• Reminds me of the Axiom of Archimedes. – marty cohen Jan 8 '19 at 0:49

This is a standard application of Baire Category Theorem (BCT). Since $$(0,\infty)=\cup_n \{x:|f(\frac x k)| \leq\epsilon \,\forall k \leq n\}$$ there exists $$n$$ such that $$\{x:|f(\frac x k)| \leq \epsilon \, \forall k \geq n\}$$ contains some open interval $$(a,b)$$. [ Because $$(0,\infty)$$, being an open subset of $$\mathbb R$$ has an equivalent complete metric so BCT applies]. This gives UNIFORM convergence of $$f(\frac x n)$$ to $$0$$ on some open interval from which the result follows easily. [ Details: let $$0<\delta and $$\delta < \frac a n$$. Then, for any $$x <\delta$$ the interval $$(\frac a x,\frac b x)$$ has length exceeding $$1$$, so it contains an interger $$k$$. Hence $$kx \in (a,b)$$. Also, $$k >\frac a x > \frac a {\delta} > n$$ so $$|f(x)|=|f(\frac {kx} k)| <\epsilon$$].

• Shouldn’t it be $\bigcup_n\{x,\,\forall k\geq n,\, |f(x/k)| \leq \epsilon\}$? – Mindlack Jan 8 '19 at 0:54
• @Mindlack You are right. That was a silly mistake I made. – Kavi Rama Murthy Jan 8 '19 at 5:01