# Thread: [SOLVED] A weird limit proof

1. ## [SOLVED] A weird limit proof

Hi, how would you solve the following problem

Prove: If $f(x)$ is defined on $\Re$ and continuous at $x=0$, and if $f(x_{1}+x_{2})=f(x_{1})+f(x_{2})$ $\forall x_{1},x_{2} \in \Re$, then $f(x)$ is continuous at all $x \in \Re$

By the way, is there a theorem that clearly states that any real number can be expressed as the sum of two other real numbers? I'm guessing this is a big part of the proof.

**EDIT**
Corrected the problem

2. Originally Posted by akolman
Hi, how would you solve the following problem

Prove: If $f(x)$ is defined on $\Re$ and continuous at $x=0$, and if $f(x)=f(x_{1})+f(x_{2})$ $\forall x_{1},x_{2} \in \Re$, then $f(x)$ is continuous at all $x \in \Re$

By the way, is there a theorem that clearly states that any real number can be expressed as the sum of two other real numbers? I'm guessing this is a big part of the proof.
Do you mean to say $f(x_1+x_2) = f(x_1)+f(x_2)$?

3. Originally Posted by ThePerfectHacker
Do you mean to say $f(x_1+x_2) = f(x_1)+f(x_2)$?
Yes. I think I now get the problem, this changes everything.

4. $f(x+h)=f(x)+f(h)$

Taking the limit: $\lim_{h\rightarrow{0}}f(x+h)=f(x)+\lim_{h\rightarr ow{0}}f(h)$ (1)

$f(0+0)=f(0)+f(0)$ thus $f(0)=0$

And now using the continuity around x=0 in (1) we get our result