I encountered a proof question:
Suppose $f(x)$ is uniformly continuous in $[0,+\infty)$. For $\forall$ x $\in$ [0,1], $\lim\limits_{n\to+\infty} f(x+n)=0$ ($n\in \mathbb{N}$). Prove that $\lim\limits_{x\to+\infty}f(x)=0$.
The question is whether the uniform continuity of $f(x)$ is necessary. When I use $x-[x]$ to represent $x\in [0,1]$, use $[x]$ to represent $n$, then $n\to +\infty$ means $[x] \to +\infty$, which can deduce $x\to+\infty$, then complete the proof. But condition uniform continuity of $f(x)$ is not used. Can someone point out the loopholes of my proof?
It is possible to find a counterexample which shows that even if $f$ is continuous, and, for all $x\in[0,1]$, $\lim_{n\to\infty}f(x+n)=0$, this is not enough to conclude $\lim_{x\to\infty}f(x)= 0$ (hence, you need something stronger than continuity to conclude). Consider a function $f$ defined as follows: For $n\ge 2$, let $I_n=[1/n+n-1/n^2, 1/n+n+1/n^2]$, and let $f$ be a continous function s.t.: $$ f(x)= \begin{cases} 0 & x\notin \bigcup_n I_n\\ >0 & x\in \bigcup_n I_n\\ 1 & x\in \bigcup_n \{1/n +n\} \end{cases} $$
This function doesn't converge to 0. However:
For $x=0, x+n\notin I_n=[(1/n-1/n^2)+n, (1/n+1/n^2)+n]$ for any $n$, thus $f(x+n)=0$ for all $n$. Same for $x > 3/4$.
For $x\in (0, 3/4]$, there will be some instant $n_0$ s.t. for all $n\ge n_0$, $x>1/n + 1/n^2$, thus, $x+n\notin I_n=[(1/n-1/n^2)+n, (1/n+1/n^2)+n]$ for all $n\ge n_0$. Hence, $f(x+n)=0$ for all $n\ge n_0$
In short, for all $x\in[0,1]$, $\lim_{n\to\infty}f(x+n)=0$.
As for errors in your proof. You say that, if you take a sufficiently large $x$, then $x=(x-[x]) + [x]$ with $[x]$ large, and because $\lim_{n\to\infty}f(x-[x]+n)=0$, $f(x)$ will be small. The problem I'd say is that for some such $x$, the sequence $(f(x-[x]+n))_n$ may fall to zero much later than at the instant $[x]$