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Let $f:(0,\infty)\to\mathbb R$ be a continuous function. Suppose that $f$ satisfies $$\lim_{x\to \infty}\left( f(x+1)-f(x)\right) = 0.$$ Show that $\lim_{x\to \infty }\dfrac{f(x)}{x} = 0.$

I tried to use epsilon-delta argument, however it is not appropriate I think. It seems $f$ is approximately homogeneous of order $0$. But it is just my intuition and I cannot explain the details. Please help me with any comments. Thank you

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An $\epsilon$-$\delta$ argument is fine, your intiution givse you the idea: $f$ is "asymptotically periodic of period $1$" and hence should be bounded, as it is continuous. A bounded function has the limit property we want.

Now to the actual argument: Let $\epsilon > 0$. Choose $M$ such that for $x \ge M$ we have that $|f(x+1) - f(x)| < \epsilon$. As $f$ is continuous, $f$ is bounded on $[M, M+1]$, by $L$ say. Induction gives us that $$ |f(x+n) - f(x)| < n\epsilon, \qquad n \in \mathbf Z^+, x \ge M $$ and therefore $$ |f(x)| \le L + n\epsilon, \qquad x \in [M+n, M+n+1] $$ That is, we have $$ \def\a#1{\left|#1\right|}\a{\frac{f(x)}x} \le \frac{L + n\epsilon}{M+n} \le \frac Ln + \epsilon, \quad x \in [M+n, M+n+1] $$ Now choose $N$ such that $\frac Ln < \epsilon$ for $n \ge N$, and we have $$ \a{\frac{f(x)}x} < 2\epsilon, \qquad x \ge M+N $$

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  • $\begingroup$ genius...; nice to see it! Thank you. $\endgroup$ – Megadeth Oct 2 '15 at 13:33
  • $\begingroup$ this is really splendid $\endgroup$ – user300 Oct 2 '15 at 13:36
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Just a little complement to martini's excellent answer:

The function $f$ is approximately periodic with period $1$ when $x$ is large, that is, if $x \ge M$, then $|f(x+1)-f(x)|\le \epsilon$. Consequently, we expect the behavior of $f$ in $[M,M+1]$ to (approximately) repeat itself in the intervals $$I_1=[M+1,M+2],$$ $$I_2=[M+2,M+3],$$ etc.

Any $x\ge M$ must lie in one of those intervals, say $I_n$, so you can write $x = u+n$ for $u$ in the original interval $[M,M+1]$. Thus, you should be able to approximate $f(x) = f(u+n)$ by the value of $f(u)$.

Now, $f(u+n) = f(u) + [f(u+1)-f(u)] + \dots + [f(u+n) - f(u+n-1)]$, so you can apply the triangle inequality $n$ times to obtain the inequality that martini provided: \begin{align} |f(u+n)| &\le |f(u)| + |f(u+1)-f(u)| + \dots + |f(u+n) - f(u)| \\ &\le L + \epsilon + \cdots + \epsilon = L+n\epsilon\end{align}

Following this idea, and replacing $x$ back in place of $u+n$, you get $$\left| \frac{f(x)}{x} \right| \le \frac{L+n\epsilon}{M+n} $$

And his conclusion follows.

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