TITLE: Prove that function is continuous without knowing the function explicitly
QUESTION [15 upvotes]: Let $f\colon \mathbb R^+\to\mathbb R$ be a function that satisfies the following conditions:
$$\tag1 \lim_{x\to 1}f(x)=0 $$
$$\tag2f(x_1)+f(x_2)=f(x_1x_2)$$
Show that $f$ is continuous in its domain.
I managed to show that $f$ is continuous at $x=1$, but I have no idea how to continue from there. Here's what I've done so far:
Because $\lim_{x\to 1}f(x)=0$, for every ϵ>0 there exists a δ>0 so that
$$0<|x - 1|<δ⇒|f(x)-0|<ϵ$$ 
To prove continuity at $x=1$ it's enough to show that $f(1)=0$ using the condition 2):
$$f(1)+f(1)=f(1 ·1)$$
$$f(1)=f(1)-f(1)$$
$$f(1)=0$$
So now we have the definition of continuity at $x=1$:
$$|x - 1|<δ⇒|f(x)-f(1)|<ϵ$$

REPLY [34 votes]: You have shown continuity at $x=1$, i.e., $$\lim_{x\to1}f(x)=f(1).$$
Consequently, for any $x_0\ne0$
$$\lim_{x\to x_0}f(x)=\lim_{x\to x_0}f\left(\frac x{x_0}\cdot x_0\right)=\lim_{x\to x_0}\left(f\left(\frac x{x_0}\right)+f(x_0)\right)=f(1)+f(x_0)=f(x_0).$$