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Let $f$ be uniformly continuous on $(0,\infty)$, and for each $h>0$, $$\lim_{n\to\infty}f(nh)$$ exists. Prove that $$\lim_{x\to\infty}f(x)$$$ exists.

My proof is as follows: By uniform continuity $$\forall\ \epsilon>0,\ \exists\ \delta>0,\ 0<x,x'<\infty, |x-x'|< \delta, |f(x)-f(x')|<\cfrac{\epsilon}{2}. $$ And by $\lim_{n\to\infty}f(n\delta)\equiv A$ exists, we know for such a $\epsilon>0$, $$\exists\ N,\ n\geq N, |f(n\delta)-A|<\cfrac{\epsilon}{2}.$$ Thus for $\forall\ x\geq N\delta$, $$ \exists\ n\geq N, n\delta\leq x<(n+1)\delta, 0\leq x-n\delta<\delta. $$ We then have $$|f(x)-A| =|f(x)-f(n\delta)|+|f(n\delta)-A|<\cfrac{\epsilon}{2}+\cfrac{\epsilon}{2}=\epsilon.$$ We conclude that $f(x)\to A$ as $x\to\infty$.

But my question is that this $A$ clearly depends on $\delta=\delta(\epsilon)$, so that $A=A(\epsilon)$. We need to check that $A$ is an absolute number before this proof. But how can I do this? Than you.

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IMHO the proof is not correct or at least not complete. 1. You did not prove that starting from certain $X$ $|f(x)-A| < eps$ for $x>X$. 2. You did not use the fact that for any $h$ your limit exists. 3. You did not claim that the limit coincide for arbitrary $h$. It should prove too. May be I missed something and you are correct. –  Alexander Vigodner Jun 16 '14 at 2:34

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