$f(x+1)=f(x)+1 \Rightarrow \displaystyle \lim_{x\to \infty}\frac{f(x)}x=1 ?$ Let $f:\mathbb{R} \to \mathbb{R}$ be a continous function such that $f(x+1)=f(x)+1 $   for all $x\in \mathbb{R}$ . Then which of the following statements is necessarily false ?.
$(1)\displaystyle\lim_{x\to \infty} \frac{f(x)} {x^{1+\epsilon}}=0$ for all $\epsilon \gt 0$
$(2)\displaystyle\lim_{x\to \infty} \frac{f(x)} {x}$ does not exist .
$(3)
\displaystyle\lim_{x\to \infty} \frac{f(x)} {x}=1$
$(4)\displaystyle\lim_{x\to \infty} \frac{f(x)} {x^{1-\epsilon}}=+\infty$ for all $\epsilon \gt 0$
My attempt:
$f(x)=x$ satisfies the hypothesis and makes the result   in $(1), (3) $ and $(4)$ true and so they are not necessarily false .
I need to prove $(2)$ is necessarily false i.e the limit does exist.
It can be easily proved by induction
$f(x+n)=f(x)+n$ for all $ n\in \mathbb{N}$ and for all $x\in \mathbb{R}$
Let $x\gt 1$ be any real number.Then
$x=\lfloor x\rfloor +\overline{ x } \quad(*)$
where the first part is greatest integer and the second is fractional part and is less than $1$ .
Then  $f(x)=f(\overline {x})+ \lfloor x \rfloor $
Now $f$ being continous on $[0,1]$ is bounded above by some $M$ on $[0,1]$ .
Hence , we have $\displaystyle \lim_{x\to \infty} \frac {f(\overline{ x })}x=0$
Thus $ \displaystyle \lim_{x\to \infty} \frac {f(x)}x= \displaystyle \lim_{x\to \infty} \frac {\lfloor x \rfloor }x$
Now again using $(*)$ , we have
$1=\displaystyle\lim_{x\to \infty} \frac {\overline{x}}x+\displaystyle\lim_{x\to \infty} \frac {\lfloor x \rfloor }x $
$\Rightarrow  
   \displaystyle\lim_{x\to \infty} \frac {\lfloor x \rfloor }x=1$ since $0\le \overline {x}\lt 1$
Thus the limit exist and is equal to $1$.
Is everything okay ? Any alternative ideas/solution will be appreciated.
Thanks for your time.

The proof is fine. Using the perhaps more common notation $\{ x \}$ for the fractional part of $x$, you showed that
$$
\frac{f(x)}{x} = \frac{f(\{ x \})}{x} + \frac{\lfloor x \rfloor}{x}
$$
where the first fraction converges to zero (because the numerator is bounded) and the second fraction converges to one (because $x-1 < \lfloor x \rfloor \le x$).
Instead of  fiddling with the integer and fractional parts of $x$ one can also argue as follows: $g(x) = f(x) - x$ is continuous and $1$-periodic, and therefore bounded. It follows that
$$
\lim_{x\to \infty}\frac{g(x)}x=0 \implies 
\lim_{x\to \infty}\frac{f(x)}x=\lim_{x\to \infty}\frac{g(x)+x}x=1 \, .
$$
