Show that $f'(0)= \lim_{\Delta x \to 0}\frac{f(\Delta x)-1}{\Delta x} = 1$ This question is related to another question I asked here.
Specifically, using the definition of $e$ I gave in that question:

There exists a unique complex function $f$ such that

*

*$f(z)$ is a single valued function $f(z) \in \mathbb{R}$ whenever $z \in \mathbb{R}$ and $f(1) = e$

*$\forall z_{1}, z_{2} \in \mathbb{C}$, $f$ satisfies $f(z_{1} + z_{2}) = f(z_{1})f(z_{2})$

*$f$ is complex differentiable for all $z \in \mathbb{C}$.


Essentially, what I'm doing in the problem linked here is trying to prove that for $f$ such a function as meets those three criteria, $f^{\prime}(0) = 1$. The $f$ I am hoping to get eventually is $f(z) = e^{z}$, but that is what I am trying to prove; I cannot use anything specifically about $e^{z}$ here, only what is provided here in our definition of the function $f$.
Eventually, I was able to show that $f^{\prime}(0) = \lim_{\Delta z \to 0} \frac{f(\Delta z)-f(0)}{\Delta z} = f(0) \lim_{\Delta x \to 0} \frac{f(\Delta x) - 1}{\Delta x}$.
But, I don't know how to go any further!
I need to show that $f'(0) = \lim_{\Delta x \to 0} \frac{f(\Delta x)-1}{\Delta x} = 1$.
Other people have posted similar questions to this on here, but they haven't been answered particularly well. I can't assume anything about $e^z$, nor can I use anything like L'Hopital's rule, or logarithms, or series expansions.
This is probably very very simple. Even an $\epsilon$-$\delta$ proof would be good, if you could write the whole thing out. I tried that route and I wasn't able to make any progress.
Please help!!
 A: Note that we have
$$f(z_1+z_2)=f(z_1)f(z_2) \tag 1$$
and $f(1)=e$.
Then, from $(1)$ we see that $f(0+1)=f(1)=f(0)f(1)\implies f(0)=1$.
Next, we analyze the derivative of $f(z)$ and find that
$$\begin{align}
f'(z)&=\lim_{\Delta z\to 0}\frac{f(z+\Delta z)-f(z)}{\Delta z}\\\\
&=\lim_{\Delta z\to 0}\frac{f(z)f(\Delta z)-f(z)}{\Delta z}\\\\
\frac{f'(z)}{f(z)}&=\lim_{\Delta z\to 0}\frac{f(\Delta z)-1}{\Delta z}\\\\
&=f'(0)\tag 1
\end{align}$$
Note that $(1)$ hold for all $z$.  In particular, it holds for real values of $z$, say $x$.  Then, we have 
$$f'(x)=f'(0)f(x)\tag 2$$
and $f(1)=e$.  
The solution to $(2)$ is $f(x)=Ce^{f'(0)x}$.  
Since $f(0)=1$, then $C=1$.  And since $f(1)=e$, $f'(0)=1$.  
And we are done!
A: I spent enough time reading this question and the linked question where you sort of try to define $e$. To be very clear none of your questions have defined $e$.
Restricting first to the case of real variables only I want to shed some light here. One of the reasons you are having trouble trying to prove $f'(0) = 1$ is because you have not given a definition of $e$. Note that the functional equation $$f(x + y) = f(x)f(y)$$ only guarantees that $f(x)$ is differentiable for all $x$ if it is differentiable at a single point $x = 0$ and moreover $f'(x) = f'(0)f(x)$. But the value of $f'(0)$ crucially depends on the value of $f(1)$. In fact we can put this dependence as a function by setting $f(1) = t$ and $f'(0) = g(t)$. It can be proved with some effort that $g(1) = 0$ and $g'(t) = 1/t$ for all $t > 0$.

Theorem: Let $f: \mathbb{R} \to \mathbb{R}$ be a function with the property that $f(x + y) = f(x)f(y)$ and let $f'(0)$ exist and let $f(1) = t\neq 0$. Then value of $f'(0)$ is a function of $t$ (say $f'(0) = g(t)$) with the following properties $$g(1) = 0, g'(t) = \frac{1}{t}$$
Proof: In the above I have set $f(1) = t$ and assumed that $t \neq 0$. This is because if $f$ vanishes at any point (say $f(a) = 0$) then by functional equation $f(x) = f(x - a)f(a) = 0$ and so the function is identically $0$. So it is essential to have $f(1) = t \neq 0$ (so that the problem is non-trivial and interesting).
Moreover from the existence of $f'(0)$ it is easily possible to deduce that $f'(x)$ exists for all $x$ and $f'(x) = f(x)f'(0)$. Hence it follows that $f(x)$ is continuous and since $f$ is non-zero at any point, it follows by intermediate value theorem that $f(x)$ is of constant sign (namely the sign of $f(1) = t$). If $f(1) = t$ is negative then $f(2) = f(1)f(1) = t^{2} > 0$ and hence $f$ changes sign. This is not possible and hence we conclude that $f(1) = t > 0$ and hence $f(x) > 0$ for all $x$. It is now clear that $f(0) = f(0)f(0)$ so that $f(0) = 1$.
Note further that the functional equation also allows us to prove that $f(x) = \{f(1)\}^{x} = t^{x}$ where $x$ is rational and this shows clear dependence of the function $f$ on parameter $t$. Hence it makes sense to change the notation a bit and replace $f$ by $f_{t}$ and thus we have the following properties of $f_{t}$: $$f_{t}(x + y) = f_{t}(x)f_{t}(y), f_{t}(1) = t, f_{t}(0) = 1, f_{t}'(x) = f_{t}'(0)f_{t}(x)\tag{1}$$ To show that the dependence of $f_{t}$ on $t$ is genuine, I show that if $s = t$ then $f_{s} = f_{t}$. Let $h(x) = f_{s}(x)/f_{t}(x)$ and clearly $h(x)$ also satisfies the equation $h(x + y) = h(x)h(y)$ and $h(1) = 1$. As we noted earlier $h(x) = \{h(1)\}^{x} = 1$ when $x$ is rational and hence by continuity $h(x) = 1$ for all $x$ and therefore $f_{s}(x) = f_{t}(x)$ for all $x$.
Further by using $h(x) = f_{st}(x)/f_{s}(x)$ we see that $h(x)$ satisfies the functional equation and $h(1) = t$ so that $h(x) = f_{t}(x)$. Thus we have the following property: $$f_{st}(x) = f_{s}(x)f_{t}(x)\tag{2}$$ for all $x$ and all positive $s, t$.
Next we consider the function $g$ which shows dependence of $f_{t}'(0) = g(t)$ on $f_{t}(1) = t > 0$. We prove that $$g(st) = g(s) + g(t)\tag{3}$$ Clearly we can see that
\begin{align}
g(st) &= f_{st}'(0) = \lim_{h \to 0}\frac{f_{st}(h) - f_{st}(0)}{h}\notag\\
&= \lim_{h \to 0}\frac{f_{s}(h)f_{t}(h) - f_{s}(0)f_{t}(0)}{h}\notag\\
&= \lim_{h \to 0}\frac{f_{s}(h)f_{t}(h) - f_{s}(h)f_{t}(0) + f_{s}(h)f_{t}(0) - f_{s}(0)f_{t}(0)}{h}\notag\\
&= \lim_{h \to 0}\frac{f_{s}(h)\{f_{t}(h) - f_{t}(0)\} + \{f_{s}(h) - f_{s}(0)\}f_{t}(0)}{h}\notag\\
&= f_{s}(0)f_{t}'(0) + f_{t}(0)f_{s}'(0)\notag\\
&= g(t) + g(s)\notag
\end{align}
Putting $s = t = 1$ we get that $g(1) = 0$. From $(3)$ we also get $g(s) = g((s/t)\cdot t) = g(s/t) + g(t)$ so that $$g(s/t) = g(s) - g(t)\tag{4}$$ We further show that $g(t) > 0$ if $t > 1$ and from $(4)$ this will establish that $g(t)$ is strictly increasing as a function of $t$. For this purpose we use the inequality $$t^{x - 1}(t - 1)\leq \frac{t^{x} - 1}{x} \leq (t - 1)\tag{5}$$ for all rational $x > 0$ and $t > 1$ (for proof of the inequality see this post : equations $(9)$ and following). Translating the above in terms of our function $f_{t}$ we get $$f_{t}(x - 1)(t - 1)\leq \frac{f_{t}(x) - f_{t}(0)}{x} \leq t - 1$$ for all $t > 1$ and all rational $x$. By continuity of $f_{t}$ the same equation holds for all real $x > 0$. Letting $x \to 0^{+}$ we get $$\frac{t - 1}{t} \leq f_{t}'(0) \leq t - 1$$ or $$\frac{t - 1}{t} \leq g(t) \leq t - 1\tag{6}$$ From this equation it follows that $g(t) > 0$ for $t > 1$ and hence $g(t)$ is strictly increasing.
Diving equation $(6)$ by $(t - 1)$ and letting $t \to 1^{+}$ we get via Squeeze theorem $$\lim_{t \to 1^{1+}}\frac{g(t)}{t - 1} = 1$$ It can be easily proved that the same limit holds when $t \to 1^{-}$ by putting $t = 1/s$ and letting $s \to 1^{+}$. Hence $$\lim_{t \to 1}\frac{g(t)}{t - 1} = \lim_{h \to 0}\frac{g(1 + h)}{h} = 1\tag{7}$$ It is now easy to prove that $g'(t) = 1/t$. Clearly
\begin{align}
g'(t) &= \lim_{h \to 0}\frac{g(t + h) - g(t)}{h}\notag\\
&= \lim_{h \to 0}\frac{g((t + h)/t)}{h}\notag\\
&= \lim_{h \to 0}\frac{g(1 + (h/t))}{(h/t)}\cdot\frac{1}{t}\notag\\
&= \frac{1}{t}\notag
\end{align}
The proof of the theorem is now complete.

We are now in a position to define $e$. Since $g(2) > 0$ and $$g(2^{n}) = ng(2)$$ it follows that $g(2^{n}) \to \infty$ as $n \to \infty$ and hence by continuity $g(t)$ takes all possible positive values. By its strictly increasing nature $g$ takes every such value only once. We define $e$ to be the unique positive number such that $g(e) = 1$. Thus if $f(1) = e$ then $f'(0) = 1$. The function $g$ is traditionally denoted by $\log$ or $\ln$. It can also be easily established that the function $g$ is inverse of $f_{e}$.
