Show that $ x\mapsto e^x$ is continuous. How can I show that $x\mapsto e^x$ is continuous ? We suppose that we don't now anything about this function except that that for all $x\in\mathbb R$ that $$e^{x+y}=e^xe^y$$
So I have that $$e^{x+h}-e^x=e^x(e^h-1)$$
but I have to show that $$\lim_{h\to 0}(e^h-1)=0.$$
How can I do it ? 
 A: You cannot prove it if your only assumption is that
$$
e^{x+y} = e^x e^y.
$$
In fact with the axiom of choice it is possible to construct an additive function which is not linear. With this additive but not linear function you are able to construct a function satisfying the property above which is not continuous in any point. 
One of the following is enough to prove the continuity:


*

*$e^x$ is continuous in at least one point

*$e^x$ is monotone

*there is a non-empty interval where $e^x$ is bounded

*$e^x$ is measurable
A: Since there exist discontinuous functions $f$ satisfying $f(x+y) = f(x)f(y)$ you'll need a bit more than just that. Since $$e^{x+y} - e^x = e^x(e^y - 1)$$ it would suffice to know that $g(x) = e^x$ is continuous at the single point $x=0$.
A: Well, continuity is a consequence of convexity, but the strategy for proving that $f(x)=e^x$ is a convex function depends on the way such function is defined. Ultimately, it boils down to proving that:
$$ \lim_{h\to 0}\frac{e^h-1}{h}=1,$$
giving $\frac{d^2}{dx^2}e^x=\frac{d}{dx}e^x = e^x$ as a consequence.
