TITLE: Proof that if $f(x+y) = f(x)f(y)$, then either the limit as $x$ approaches $0$ is $1$, or $f(x)=0$
QUESTION [5 upvotes]: Assume that $f: \mathbb{R} \to \mathbb{R}$ is such that $f(x+y)=f(x)f(y)$ for all $x, y \in \mathbb{R}$. If $f$ has a limit at zero, prove that $f$ has a limit at every point and either $\lim_{x \to 0}f(x)=1$ or $f(x)=0$ for all $x \in \mathbb{R}$.
Thanks in advance.

REPLY [2 votes]: I add this little detail just because the question asks for it: given $x \in \mathbb{R}$ and $h \neq 0$, we have $f(x+h)-f(x)=f(x)f(h)-f(x)=f(x)(1-f(h)) \rightarrow 0$ when $h \rightarrow 0$ as the limit at zero exists and is 1. This proves that the limit exists at every point.