Limit of a function satisfying an inequality If $f(x)+f(y)\leq f(x+y)$ and $f:\mathbb{R}\to\mathbb{R}$, then can we find $\lim_{x\to 0} \frac {f(x)}{x}$?
I am not sure whether the question is correct.Thank you.(I tried this idea: $f(x)=f(x+y-y)\ge f(x+y)+f(-y)\ge f(x)+f(y)+f(-y)\implies f(y)\leq -f(-y)$ but after that I seem to be hitting a roadblock.
Thank you in advance.
 A: A function $f:{\mathbb R}\to{\mathbb R}$ which fulfills
$$(\forall x,y\in\mathbb R)f(x+y) \le f(x)+f(y)$$
is called is subadditive.
It was already mentioned in comments that the limit need not exist without any additional assumptions on $f$. E.g. if $f$ is any non-linear solution of Cauchy's equation, then the limits does not exists, but the function is both subadditive and superadditive.
The following result from the book An introduction to the theory of functional equations and inequalities By Marek Kuczma p.467 gives at least some conditions when the limit exists:

Theorem 16.3.3. Let $f:\mathbb R\to\mathbb R$ be a measurable subadditive function, and let
  $$A = \inf_{t<0} \frac{f(t)}t, \qquad B=\sup_{t>0} \frac{f(t)}t.$$
  If $A$ resp. $B$ is finite, then
  $$A = \lim_{h\to0^-} \frac {f(h)}h,\text{ resp. }B=\lim_{h\to0^+} \frac {f(h)}h.$$
  The above formulas remain valid for $A$ and/or $B$ infinite under the additional assumption
  that $\lim\limits_{x\to 0} f(x) = 0$, or $\liminf\limits_{x\to 0} f(x)>0$
  Moreover, in every case,
  $$A \le B.$$

If you rewrite the above results for the function $g(x)=-f(x)$, you get results for superadditive functions, i.e. $g(x+y)\ge g(x)+g(y)$.
