# 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.

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Are you assuming continuity or anything? There exist discontinuous $f$ such that $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb R$, in for those, the limit will not exist. –  Jonas Meyer Jan 21 '12 at 6:13
To add to @Jonas' remark: Every function of the form $f(x) = ax$ (with $a$ real) satisfies the inequation. For such a function, the limit exists and is equal to $a$. So it is possible that the limit exists but cannot be found from the given data alone. Can you clarify what you really want to show? –  Srivatsan Jan 21 '12 at 6:18
@Rajesh: Srivatsan's point is that even among functions for which the limit exists, it is not uniquely determined from the hypotheses. There exist $f$ where the limit is $0$, $46$, or $-\sqrt 2$, and others for which it doesn't exist. –  Jonas Meyer Jan 21 '12 at 6:35
@Jonas: However, I think it pretty clear that the OP wants to ask whether the hypothesis implies that the limit exists, not whether the value of the limit can be inferred from the hypothesis alone. –  Brian M. Scott Jan 21 '12 at 6:48
@Brian: Your guess may well be right, but it is not clear to me. "Can we find" the limit often means more, namely finding the value of the limit. Hopefully Sabyasachi Mukherjee will clarify. –  Jonas Meyer Jan 21 '12 at 7:29

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.
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. $g(x+y)\ge g(x)+g(y)$.