Prove that $\lim_{h\to 0}\frac{e^h-1}{h}=1$ from the functional equation $f(x+y)=f(x)f(y)$.

Question

If I have that $e^x$ is defined by the unique continuous function $\mathbb R\to \mathbb R^*$ s.t. $f(x+y)=f(x)f(y)$ for all $x,y\in\mathbb R$ and $f(1)=e$, is it possible to prove that $$\lim_{h\to 0}\frac{e^h-1}{h}=1\ \ ?$$

Answer 1

Note that $$ \left(1+\frac1n\right)^{n}=\sum_{k=0}^n {n\choose k}\frac1{n^k}=\sum_{k=0}^n \frac1{k!}\,\prod_{j=1}^{k-1}\left(1-\frac jn\right). $$ From this one can show that $$ e=\sum_{k=0}^\infty \frac1{k!}. $$ Now define $$g(x)=\sum_{k=0}^\infty \frac{x^k}{k!}.$$ This converges uniformly on any bounded set, so $g$ is in particular continuous. It is also differentiable with $g'=g$ and $g(1)=e$. Fix $y$. Then $h(x)=g(x+y)$ satisfies $h'(x)=h(x)$, $h(0)=g(y)$. This is a solution to the initial value problem $h'=h$, $h(0)=g(y)$, of which $g(y)g(x)$ is also a solution. By the uniqueness of the solution (Picard) we get that $g(x+y)=g(x)g(y)$. So $g$ is a continuos function with $g(x+y)=g(x)g(y)$ and $g(1)=e$. By your hypothesis, $g=f$. As $g$ is differentiable, $$ \lim_{h\to0}\frac{e^h-1}h=\lim_{h\to0}\frac{f(h)-1}h=\lim_{h\to0}\frac{g(h)-g(0)}h=g'(0)=1. $$