Limit as $x$ goes to zero of an arbitrary function

Let $f:\mathbb R\to\mathbb R$ be a function with the property that $f(a+b)=f(a)+f(b)$ for all real numbers $a$ and $b$. Assume that the limit as $x\to 0$ of $f(x)$ is equal to some real number $L$. Show $L=0$.

Edit: While I know that the functions for which this is true are, for example, $f(x)=cx$, I can't assume anything that is not given.

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Is $f$ continuous? – kneidell Dec 13 '12 at 18:17
@kneidell I don't think you need continuity, just that $\lim_{x\to 0} f(x)$ exists. There are definitely functions of this sort which are not continous on all of $\mathbb R$, but are continuous at $0$. – Thomas Andrews Dec 13 '12 at 18:19
(I should have said, assuming axiom of choice, there are definitely functions of this sort... Basically, you just need a basis of $\mathbb R$ as a vector space over $\mathbb Q$ to get non-continuous examples. Indeed, you don't need even a basis, just a representation of $\mathbb R\cong V\oplus W$ where $V$ and $W$ are two non-trivial $\mathbb Q$-vector spaces. – Thomas Andrews Dec 13 '12 at 18:46

Let $a=b=0$. Then $$f(0)=f(0)+f(0).$$ So we know $f(0)$.

Let $a=x$, $b=-x$. Then $$f(a+b)=f(0)=f(x)+f(-x).$$ Let $x\to 0$.

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Thank you! I think I understand the question better now – Hayley Dec 13 '12 at 18:29

Note $$\tag{1}f(a+a)=2f(a).$$ Also note that as $a\rightarrow 0$, we also have $a+a\rightarrow 0$. From $(1)$, we deduce that $L=2L$; whence $L=0$.

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Thank you! That helps a lot. – Hayley Dec 13 '12 at 18:29

Hint: If $f$ is identically zero, you are done. Otherwise, find some $x$ for which $f(x) \neq 0$ and consider $f(nx)$ for large $n \in \mathbb{N}$.

Edit: Ah, misread the problem. I thought you were interested in the limit as $x \rightarrow \infty$. I'll leave this up, just for fun.

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