A form of the Cauchy functional equation 
Find all the functions $f :\mathbb R \to \mathbb R$ that satisfy the
conditions:

*

*$$f(x+y)=f(x)+f(y), \enspace \forall x,y \in \mathbb R;$$

*$$\exists \lim_{x\to \infty}f(x).$$

This problem is important for the community because on the forum I only saw the case where $f$ is continuous, but not the case when we only know that $\displaystyle\exists \lim_{x\to \infty}f(x).$
I found a solution to this problem in a book of analysis I own, which hints the following:
It is easy to see that $f(q)=q\cdot f(1), \enspace \forall q\in \mathbb Q$. Let $f(1)=a$.
If $a>0$, then consider the sequence $a_n=n.$ It follows that $$\lim_{x\to \infty}f(x)=\lim_{n\to \infty}f(a_n)=\lim_{n\to \infty}an=\infty.$$
The next step in the book is that this implies $f$ is increasing. I couldn't understand this line properly. Why is $f$ increasing? Of course, it is easy to check that $f(q)\le f(r), \forall q<r \in \mathbb Q$, but why is this also true for reals?
Please help me understand this! Thank you so much!
It is also clear that if $f$ is increasing then $f(x)=ax$ for all $x\in \mathbb R$ (suppose there exists an $x\in \mathbb R\setminus \mathbb Q$ such that $f(x)<x$. Then by density there is a $q\in \mathbb Q$ such that $f(x)<aq=f(q)<ax$, and this clearly gives the contradiction that $x< q <x$).
 A: So we have that lim $f = +\infty $.
Let's $x<y$ be real numbers. Then $f(y)-f(x) = f(y-x)$, so $n(f(y)-f(x)) = f(n(y-x))$.
Since $f(n(y-x)) \rightarrow +\infty$ when $n\rightarrow +\infty$, it is positive for $n$ big enough.
So $n(f(y)-f(x))$ is positive for $n$ big enough. But the sign of $n(f(y)-f(x))$ does not depend on $n$.
So $n(f(y)-f(x))$ is always positive, and especially $f(y) > f(x)$.
A: I do not think you need the increasing property and can argue more directly:
As before, $f(qx)=qf(x)$ for $q\in \Bbb Q$ and $x \in\Bbb R$, and let $a=f(1)$
Let $y\notin\Bbb Q$. Assume $f(y)\ne ay$, say $\epsilon:=\left|\frac{f(y)}y-a\right|>0$. By density of $\Bbb Q$ in $\Bbb R$, we find $u,v\in \Bbb Q$ with $y-\epsilon<u<y<v<y+\epsilon$.
Then $|f(y)-ay|>|a|\epsilon$ whereas $|f(u)-ay|$ and $|f(v)-ay|$ are both $<|a|\epsilon$. So
$f(u)$, $f(v)$ are either both $>f(y)$ or both $<f(y)$. At any rate, we find two positive numbers $x_1=y-u$ and $x_2=v-y$ such that exactly one of $f(x_1)$, $f(x_2)$ is positive and the other negative. Then one of $\lim_{n\to\infty}f(nx_1)$, $\lim_{n\to\infty}f(nx_2)$ is $+\infty$ and the other is $-\infty$. This contradicts the existence of $\lim_{x\to\infty}f(x)$.
