Show that unifomrly continuous function of half closed interval have limit. unifomrly continuous function 
$$f: [0, \infty) \rightarrow \mathbb{R}$$
if f is satisfying 
$$\lim_{n\rightarrow +\infty}{ }f(n+x)=0, x\in\left [0,1  \right ]$$
show that $$\lim_{x\rightarrow +\infty}{ }f(x)=0$$
sol) first all , I supposed to $$\lim_{n\rightarrow +\infty}{ }x_{n}=0$$
so,I think that show
$$\lim_{n\rightarrow +\infty}{ }f(n+x)=0=\lim_{n\rightarrow +\infty}{ }x_{n}$$
but, I know that this statement also satisfying f is continuous function.
I should be show that 
if f is uniformly continuous function $$\lim_{x\rightarrow +\infty}{ }f(x)=0$$
 if f is continuous function  $$\lim_{x\rightarrow +\infty}{ }f(x)\neq0$$
I need your help. And thank you for read this question.
 A: You're welcome! Let's show for any $\epsilon > 0$ that $$ - \epsilon < \liminf_{x \rightarrow \infty} f(x) \leqslant \limsup_{x \rightarrow \infty} f(x) < \epsilon$$
Well, for any $\epsilon'$, we get a $\delta'$ such that if any two points are no more than $\delta'$ apart, $f$ of them is no more than $\epsilon'$ apart. WLOG $\delta' = \frac{1}{N}$ for some $N \in \mathbb{N}$. 
Now let's sample at $\mathbb{N}, \mathbb{N} + \frac{1}{N}, \ldots \mathbb{N} + \frac{N-1}{N}$ (draw a picture!). Because there are finitely many sequences, for $M \in \mathbb{N}$ large enough $\lvert f(M) \rvert, \ldots, \lvert f(M + \frac{N-1}{N}) \rvert < \epsilon''$. By uniform continuity,$\lvert f(M + x) \rvert < \epsilon'' + \epsilon'$, for any $x \in [0,1]$.  
Edit: for an example of $f$ just continuous, for which the conclusion $\lim_{x \rightarrow \infty} f(x) = 0$ fails, the following works: on each interval $[N, N+1)$, for $N \in \mathbb{N}$, take a bump function supported on $(N+x_N, N+y_N)$, where $0 < x_N < y_N < 1$, $\lim_N x_N = \lim_N y_N = 1$, i.e. the bump functions are squeezed onto the right side of the interval as the time $N$ increases. You can check the resulting function has the desired property. 
