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Let $f:(a,\infty)\rightarrow \mathbb{R}$ be a function such that $f$ is bounded in any finite interval $(a,b]$. Prove that $\lim_{x \to \infty} f(x)/x=\lim_{x \to \infty}[f(x+1)-f(x)]$, provided that the right limit exists. Thanks a lot.

JavaMan
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S Ali Mousavi
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    Maybe you are missing a hypothesis. Let $f(x)=\sin{x} +1$. The left hand side is zero and the right hand side is not defined. On the other hand if you add the hypothesis that $\lim_{x\rightarrow \infty}f’(x)$ exists, it’s true. – Charlie Frohman Dec 20 '18 at 19:20
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    That requires differentiability, and the question doesn't even require continuity. Might there be a slightly better approach, perhaps $f$ is strictly monotonic? @CharlieFrohman I suppose then, we'd have the bounded part by default. – Thomas Andrews Dec 20 '18 at 19:24
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    Now, what have you tried? – Thomas Andrews Dec 20 '18 at 19:30
  • Thanks. I edit my question. My attempt is to write $f(x+1)-f(x)=L+e(x)$ such that $\lim_{x \to \infty} e(x)=0$. But i could not prove it. – S Ali Mousavi Dec 20 '18 at 19:32

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Let $c=\lim f(x+y)-f(x)$. Given $\epsilon>0$, there is $M$ such that $|f(x+y)-f(x)-c|<\epsilon$ for $x>M$. Also, $|f(x)| < L$ for $x\le M+1$. Then for $x>M$ and with $n=\lfloor x-M\rfloor$, $$-L+n(c-\epsilon)<f(x-n)+n(c-\epsilon)\le f(x)\le f(x-n)+n(c+\epsilon)<L+n(c+\epsilon) $$ and so $$ \frac {L-M(|c|+\epsilon)}x+c-\epsilon<\frac{f(x)}x<\frac {L+M(|c|+\epsilon)}x+c+\epsilon$$ which implies $$ c-\epsilon\le \liminf \frac{f(x)}x, \quad\limsup \frac{f(x)}x\le c+\epsilon.$$ As $\epsilon$ was arbitrary, $$ \lim\frac{f(x)}x=c.$$

Hagen von Eitzen
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