Limit of a Lipschitz function We have the following  lemma:
Let $f: \mathbb R \to \mathbb R$ be a Lipschitz continuous function, then:
$$\lim_{x \to \infty } f(x) = 0 \iff \forall t \in[0,1] \ \ \lim_{n \to \infty } f(t+n) = 0 \ $$
Note that if we remove the Lipschitz condition, this becomes false. (Because we can construct a function made up of arbitrarilly thin "waves"  which satisfies the second condition, but does not go to $0$.)
I'm having trouble with the "only if" direction. I tried the following. Let $ \lambda$ be the Lipschitz constant of $f$. Let $\varepsilon  > 0.$ Now we have for $t\in[0,1]$
$$ |f(t)|\leq |f(t)-f(t+n)|+|f(t+n)|$$
and for large $n$ 
$$|f(t)|\leq \lambda n+\varepsilon$$
Which is not a good bound. How can I get a better one?
 A: The idea is to use compactness to show that $f(n+t) \to 0$ "uniformly" for $t\in [0,1]$. Here is the detailed argument:
Assume that $\lim_{n\to \infty} f(t+n) = 0$ for all $t \in [0,1]$. Let $\epsilon> 0$. Then for each $t \in [0,1]$, there is $N_t \in \mathbb N$ such that $n \ge N_t$ gives $\lvert f(t + n) \rvert \le \epsilon/2$. Take $\delta$ such that $0 < \delta < \epsilon/(2\lambda)$ and take $t_{*} \in [0,1]$ such that $\lvert t - t_0 \rvert < \delta$. Then for $n\ge N_t$, we see $$\lvert f(t_* + n) \rvert \le \lvert f(t_* + n) - f(t + n) \rvert + \lvert f(t+n) \rvert \le \lambda \lvert t- t_0\rvert + \epsilon/2 < \epsilon.$$ That is, for each $t \in [0,1]$, there is $N_t \in \mathbb N$ such that $n\ge N_t$ gives $\lvert f(t_* + n) \rvert < \epsilon$ for all $t_* \in (t - \delta, t+ \delta)$. The neighborhoods $\{(t- \delta, t+ \delta)\}_{t \in [0,1]}$ form an open cover of $[0,1]$. Since $[0,1]$ is compact, there is a finite subcover: $\{(t_k-\delta, t_k + \delta)\}_{k=1}^K$. These values $t_k$ come with corresponding $N_{t_k} \in \mathbb N$ satisfying the above. Take $N = \max_{1\le k \le K}\{ N_{t_k}\}$. Then for all $n \ge N$, we have $\lvert f(t_*+n) \rvert < \epsilon$ for all $t_* \in [0,1]$. 
Now for $x \ge N$, we have $x = n_x + t_x$ for some natural number $n_x \ge N$ and $t_x \in [0,1]$ (these are simply the floor and fractional part of $x$ respectively). Then $$\lvert f(x) \rvert = \lvert f(n_x + t_x) \rvert < \epsilon.$$ That is, $x \ge N$ gives that $\lvert f(x) \rvert < \epsilon$. This shows that $\lim_{x\to\infty} f(x) = 0$.
