Prove that $\lim\limits_{t \to \infty} x(t)$ exists and is an integer. 
Let $f : \Bbb R \longrightarrow \Bbb R$ be a $\operatorname {C}^{\infty}$-function such that $f(x) = 0$ if and only if $x \in \Bbb Z.$ Suppose the function $x : \Bbb R \longrightarrow \Bbb R$ satisfies $x'(t) = f(x(t)),$ for all $t \in \Bbb R.$ If $\Bbb Z \cap \{x(t)\ |\ t \in \Bbb R \} = \varnothing,$ then show that $\lim\limits_{t \to \infty}  x(t)$ exists and is an integer.

What I have seen is that $x$ is also a $\operatorname {C}^{\infty}$- function and since image of $x$ doesn't contain any integer so by IVP it follows that $$\{x(t)\ |\ t \in \Bbb R \} \subseteq (a,b)$$ where $a, b \in \Bbb R$ with $b-a \leq 1$ such that $(a, b) \cap \Bbb Z = \varnothing.$ In other words the image of $x$ is strictly lying between two consecutive integers. Hence $x$ is bounded and also by the definition of $x'$ and $f$ it follows that $x'(t) \neq 0,$ for all $t \in \Bbb R.$
Now how do I proceed? Any help will be highly appreciated.
Thanks in advance.
 A: As you noticed, the sign of $x^\prime(t)$ is constant. Suppose that it is positive, the case negative can be dealt with in a similar manner.
As $x^\prime(t)>0$, $x(t)$ is increasing. As you noticed, $x(t)$ is also bounded. So $\lim\limits_{t \to \infty} x(t)$ exists and $\lim\limits_{t \to \infty} x^\prime(t) =\lim\limits_{t \to \infty} f(x(t))=l$ would also exist.
$l=0$ as if not suppose for example $l>0$, we would have $\lim\limits_{t \to \infty} x(t) = \pm \infty$ as according to Mean Value Theorem for $x$ large enough $f(x) \ge \frac{l}{2} x$.
Based on the fact that $f$ only vanishes on $\mathbb Z$, you can conclude that $\lim\limits_{t \to \infty} x(t)$ is an integer.
A: As noted $f$ has a sign in the domain of interest  let's say $f$ is positive here (using the $a$ from the OP, we are saying without loss of generality that $f$ is positive on $(\lfloor a \rfloor, \lfloor a \rfloor,+1)$), then $x$ is increasing and therefore (by the assumed boundedness of $x$) $l=\lim_{t\to\infty} x(t)$ exists as a real number (*).
Suppose $l\not\in \mathbb Z$. Thus $f(l)\neq 0$. There is some $\epsilon$ such that if $x\in\mathbb R$ is such that $l\ge x>l-\epsilon$, then $|f(x)-f(l)|<f(l)/2$, and hence $|f(x)|=f(x) \ge f(l)/2$.
For all sufficiently large times, we have $l\ge x(t) > l - \epsilon_1 $, and therefore  $$x'(t) = f(x(t)) >\frac{f(l)}2$$
so $x(t)\ge f(l)t/2 + C$ for some $C\in\mathbb R$, and $x(t)\to\infty$, contrary to (*).
