# Prove that $f$ having a limit at $0$ implies that $f$ has a limit at every real number

Assume $f: \mathbb{R} \to \mathbb{R}$ is such that $f(x+y)=f(x)f(y)$ ( The class of exponential functions has this property). Prove that $f$ having a limit at $0$ implies that $f$ has a limit at every real number and is one, or $f$ is identically $0$ for every $x \in \mathbb{R}$

This is what I have so far. I know we can demonstrate this using the exponential function where $$f(0)=a^0=1$$ Hence $$f(x+0)=f(x)f(0)=a^x*a^0=a^x=f(x)$$ If $f$ is continuous at $0$ then $$\lim_{h\to 0}f(x+h)=\lim_{h\to 0}f(x)f(h)=f(x)$$ Making $f$ continuous.

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Exactly the same. Forget the $a$ –  leo Feb 1 '13 at 4:05
What? I don't follow what you are trying to say –  math101 Feb 1 '13 at 4:08
Like how would I continue this? –  math101 Feb 1 '13 at 4:09
You don't actually have to assume it's an exponential function, so you can remove all the parts that use $a^x$. –  Robert Mastragostino Feb 1 '13 at 4:09
Ohhh I get it. Thanks :) –  math101 Feb 1 '13 at 4:10

For the second, set $y=0$ and get $\forall xf(x)=f(x)f(0)$. If there is any number $x$ such that $f(x) \ne 0$ we get that $f(0)=1$
What does it mean that f is identically $0$ for $x\in \mathbb{R}$? –  math101 Feb 1 '13 at 4:50
@math101: that is a perfectly good function. No matter what $x$ you feed it, $f(x)=0$. In this proof, it means I can't find an $x$ so I can divide by $f(x)$, so the proof that $f(0)=1$ fails. A good thing, because $f(0)=0$. –  Ross Millikan Feb 1 '13 at 5:02