Proof that if $f(x+y) = f(x)f(y)$, then either the limit as $x$ approaches $0$ is $1$, or $f(x)=0$ 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.
 A: What does this function remind you of?  It may be a bit hard to see at first, but do we know any functions where when we plug in $x+y$ we can "split" it through multiplication?  What about $e^x$?  $e^{x+y} = e^x e^y$, and actually any exponential function $a^x$ for a constant $a \in \mathbb{R}$ will satisfy the property you need.
Let's investigate the limit of $f(x)$ as $x \to 0$.  We are assuming that the limit does indeed exist, so we can do this manipulation:
$$\lim_{x \to 0} f(x) \cdot \lim_{x \to 0} f(x) = \lim_{x \to 0} f(x)f(x) = \lim_{x \to 0} f(x + x)$$  
But as $x \to 0$, $x+x \to 0$ as well, so in essence the last part $\lim_{x \to 0} f(x + x)$ is just saying $\lim_{x \to 0} f(x)$.
Thus, we have the equality
$$\lim_{x \to 0} f(x) \cdot \lim_{x \to 0} f(x) = \lim_{x \to 0} f(x)$$ or in other words (since we know the limit exists, we can call it $L$) we have $$L^2 = L$$
Also, since as $x \to 0$, $-x \to 0$ as well we have that $$L = L^2= \lim_{x \to 0} f(x) \lim_{x \to 0}f(-x) = \lim_{x \to 0} f(x)f(-x) = \lim_{x \to 0} f(x-x) = \lim_{x \to 0} f(0) = f(0)$$
What real numbers have this property?  Only $1$ and $0$.  So now, we have that $\lim_{x \to 0} f(x) = 1$ or $\lim_{x \to 0} f(x) = 0$.  If it is $1$ then we have completed the proof, so suppose it is $0$.  Then we can write any real number $a \in \mathbb{R}$ as $a+0$ and so we have that $$f(a+0) = f(a)f(0) = f(a) \cdot 0 = 0$$
Hence, we have completed the proof.
A: 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.
