$f(x)$ is uniform continuous and $\{f(nh)\}_{n\in\mathbb N}$ converges,prove $\lim_{x\to \infty}f(x)$ exists Assume $f(x)\in C[0,+\infty)$ and $f(x)$ is uniform continous,if for any $h>0$,the sequence $\{f(nh)\}_{n\in\mathbb N}$ converges,please prove that $\lim\limits_{x\to\infty}f(x)$ exists.
I have no idea how to answer this question,though it doesn't seem to be diffcult.Please give me some ideas or solutions,thank you.
 A: Pick $h=1$, assume that the sequence $\{f(n)\}_{n=1}^\infty$ converges to $L$. Then for each $m\in \mathbb N$, let $h = 1/m$. Then $\{ f(n/m\}_{n=1}^\infty$ contains the subsequence $\{ f(n)\}_{n=1}^\infty$. Thus $ \{ f(n/m\}_{n=1}^\infty$ converges to $L$ for all $m$. 
Now we show that $\lim_{x\to \infty} f(x) = L$. 
Let $\epsilon>0$. Since $f$ is uniform continuous, there is $\delta >0$ so that if $x, y\in \mathbb R$ and $|x-y|<\delta$, then 
$$\tag{1} |f(x) - f(y)|< \epsilon/2.$$ 
Now let $m\in \mathbb N$ so that $1/m < \delta$. Since $\{ f(n/m\}_{n=1}^\infty$ converges to $L$, there is $N\in \mathbb N$ so that 
$$\tag{2} |f(n/m)-L|<\epsilon/2$$
for all $n\ge N$. Let $M = N/m$. Then if $x\ge M$, there is $n\ge N$ so that $|x-n/m|<\delta$ (we used $1/m<\delta$ here). Then $|f(x) - f(n/m)|<\epsilon/2$ and thus 
$$|f(x) - L|\le |f(x) - f(n/m)|+|f(n/m)-L|<\epsilon.$$
Since $\epsilon>0$ is arbitrary we conclude $\lim_{x\to \infty} f(x) = L$. 
A: Note that $f(n)$ converges to $L$. Further, $f(\frac{n}{2})$
converges to $L$. And $\lim_n\ f(\frac{ n}{2^k}) =L$ for any $k$. Here note that $\{ \frac{n}{2^k}\}$ is dense in $[0,\infty)$.
Whence $\lim\ f(x)=L$ : If not, there is $x_n$ s.t.
$|f(x_n)-L|>\delta>0$.
When $ |x_n- \frac{ n'}{2^k} | <\delta_2$, then
$|f(x_n)-f(\frac{n'}{2^k} )| < \delta/2$ by uniform continuity. Then
$|f(x_n) - L|\leq \delta/2$ for large $n$, since $f(\frac{n'}{2^k} )
$ goes to $L$.
