# How to prove this simple limit?

Let $f:[0,+\infty]\to R^1$ to be a function which satisfies the following properties.  1.$f$ is uniformly continuous on $[0,+\infty)$  2.For any $x_0\in[0,1]$,we always have $\lim_{n\to+\infty}f(x_0+n)=0$  Please find $\lim_{x\to+\infty}f(x)$.  It's obvious to know that the limit is zero, but my question is how to use mathematics language to prove instead of my intuition. Is there any definition or theorem that I can apply to solve this? (A hint would be grateful.)

• What does it mean by $R^1?$@Chiu-Tzu-Hsuan
– diya
Dec 19, 2015 at 12:30
• @diya it means its domain is from 0 to $\infty$ in real space with dimension 1 Dec 20, 2015 at 7:51

Fix $\varepsilon>0$. Since $f$ is uniformly continuous on $\left[0,\infty\right)$, there is some $\delta>0$ such that if $\left|x_1-x_2\right|<\delta$, we have $\left|f\left(x_1\right)-f\left(x_2\right)\right|<\frac{\varepsilon}{2}$.
Now the idea is to take $m\in\mathbb{N}$ such that $m>\frac{1}{\delta}$, and by the uniform continuity, the value of $f$ at a point $x\in\mathbb{R}$ can be "approximated" by $f\left(\frac{k}{m}+n\right)$ for appropriate $k=0,\dots,m-1$ and $n$. We only need to choose $n=\left\lfloor x\right\rfloor$ and $k$ such that $\left|\frac{k}{m}+n-x\right|<\frac{1}{m}<\delta$.
We can use the definition of limit and prove that for every $\epsilon>0$ there exists $M>0$ such that $|f(x)|<\epsilon$ when $x>M$. So let's fix $\epsilon>0$, since $f$ is uniformly continuous, there is a $\delta>0$ such that $|f(x)-f(y)<\epsilon/2$ if $|x-y|<\delta$. Now divide the interval $[0,1]$ in $m$ intervals of lenght $1/m$ so that $1/m \le \delta$ : $[0=x_0,x_1],[x_1x_2],...,[x_{m-1},x_n=1]$. By hypothesis 2) we know that for every $x_i$ there exists $M_i>0$ such that $|f(x_i+M_i)|<\epsilon/2$ for every $n>M_i$. Take $$M=\max M_i+1.$$ This works because if you choose $x>M$, then $x-\lfloor x \rfloor \in [0,1]$, in particular $x-\lfloor x \rfloor \in [x_i,x_{i+1}]$ for some $i$. Hence $|f(x_i+\lfloor x \rfloor)|<\epsilon/2$, because $\lfloor x \rfloor\ge M>M_i$, and so $$|f(x)|\le |f(x_i+\lfloor x \rfloor)|+\epsilon/2\le\epsilon$$ because $|x-(x_i+\lfloor x \rfloor)|=|x-\lfloor x \rfloor-x_i|<1/m \le \delta$.