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Let $f$ be continuous such that $$\lim _{x\to \infty }\left(f\left(x+1\right)-f\left(x\right)\right)=0$$ show that $$\lim _{x\to \infty }\left(\frac{f\left(x\right)}{x}\right)=0$$

My idea was applying the sequential criterion for continuity: that if $f$ is continuous at $c$ if and only if for every sequence $x_n$ that converges to $c$, $f(x)$ converges to $f(c)$. Also, I have the idea that since $$\lim_{x\to \infty }\left(f\left(x\right)\right)=\lim_{x\to\infty}\left(f\left(x+1\right)\right),$$ I can use it somehow in the demonstration, or am I wrong?

Botond
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Jojo98
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  • The last two limits you've written might not exist, so you cannot use something like that. – Clayton Feb 13 '18 at 20:46
  • If we had instead $f(x+1)-f(x)=0,$ then $f$ would be periodic with period $1$ and the theorem would be easy to prove. So I would start by proving that that, and then trying to modify the proof to cover the given situation. Just a suggestion -- I haven't tried it. – saulspatz Feb 13 '18 at 20:53
  • Assume that $f(x)/x$ does not approach $0$ and see if you find this assumption contradicts the hypothesis. – Mark Viola Feb 13 '18 at 21:15
  • See https://math.stackexchange.com/questions/2119937/limit-of-ratio-fx-over-x-equal-to-limit-of-difference-fx1-fx?noredirect=1&lq=1 – xpaul Feb 13 '18 at 21:40
  • @ConnorHarris In the question you pointed to the function $f$ is supposed bounded. – egreg Feb 13 '18 at 23:07
  • @egreg Bounded on finite open intervals, not on all of $\mathbb{R}$. – Connor Harris Feb 14 '18 at 14:28

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