Prove that $f$ is continuous Let $f$ be a non-constant function such that $f(x+y)=f(x)\cdot f(y),\forall x,y\in\mathbb{R}$ with $\lim_\limits{x\to 0}{f(x)}=l\in\mathbb{R}$. Prove that:
a) $\lim_\limits{x\to x_0}{f(x)}\in\mathbb{R},\forall x_0\in\mathbb{R}$
b) $l=1$
I have managed to prove a:
$$\lim_\limits{h\to 0}{f(x_0+h)}=\lim_\limits{h\to 0}{(f(x_0)\cdot f(h))}=l\cdot f(x_0)\Rightarrow \lim_\limits{x\to x_0}{f(x_0)}=l\cdot f(x_0)\in\mathbb{R}$$
But I cannot move further to b, except proving that $f(0)=1$:
For $y=0$ in the original relation, we get: $f(x)\cdot (f(0)-1)=0,\forall x\in\mathbb{R}$ and since $f$ is non-constant we get that $f(0)=1$. Any hint for b?

Assuming that you have established part a) and the fact that $f(0) = 1$ we can proceed to show that $l = 1$. Clearly we can see that $f(x)f(x) = f(2x)$ and hence taking limits when $x \to 0$ (so that $2x \to 0$) we get $l^{2} = l$. So either $l = 0$ or $l = 1$.
Again note that $f(x)f(-x) = f(0) = 1$ and taking limits as $x \to 0$ (and noting that $-x \to 0$) we get $l^{2} = 1$ so that $l = \pm 1$. Thus the only option is to have $l = 1$.
