How can I find the example of $f(x)$ such that $\,\lim_{x\to\infty}f(x) \neq 0$? 
Let $f:[0,+\infty)\longrightarrow  R^{+}\bigcup\{0\}$ be a continous and for any $x\in[0,+\infty)$ the sequence 
  $\{f(x+n)\}$ converges to zero,prove that
  $$\lim_{x\to+\infty}f(x)=0$$

I think this problem is wrong, so someone can take some example? Thank you,
meaning that
find a $f$ such:
let $f:[0,1]\longrightarrow  R^{+}\bigcup\{0\}$ be a continous and for any $x\in[0,1]$ the sequence 
$\{f(x+n)\}$ converges to zero,prove that
$$\lim_{x\to+\infty}f(x)\neq 0$$
someone tell me 
$$f(x)=\dfrac{x}{1+x^2\sin{x}}$$
But I think this is example is not such my meaning.Thank you
and I have seen this problem :

 A: I'm not sure whether your intended question includes the uniform continuity condition or not; if so then here's a proof. If not, then I described a counterexample in the comments.
Any sequence in $[0,\infty)$ can be written uniquely as $\{n_j + x_j\}$ where $n_j \in \mathbb{N}, x_j \in [0,1)$. Take any such sequence converging to $+\infty$; we must show that $f(n_j + x_j) \to 0$.
Since $[0,1]$ is compact we can pass to a subsequence (which I will refer to also as $(n_j,x_j)$ for convenience) such that $x_j \to x_0 \in [0,1]$. Then by the uniform continuity we have
$$ f(n_j + x_j) \le f(n_j + x_0) + \omega(|x_j - x_0|) $$
where $\omega$ is the modulus of continuity for $f$. The first term on the RHS converges to zero by our assumptions and the second by the definition of the modulus of continuity and $x_j \to x_0$; so we have $f(n_j + x_j) \to 0$.
This argument in fact works for any subsequence; so every subsequence has a subsequence on which $f$ converges to zero, and thus $f$ converges to zero on the original sequence; so since it was arbitrary we have shown $f(x) \to 0$ as $x\to \infty$.
A: As a counterexample take any function $f$ which is $0$ in each interval $[n+2/n,n+1]$ and is $1$ in the points $n+1/n$ for $n\in \mathbb N$. For each $x$ the sequence $f(x+n)$ will be definitely $0$.
