Please prove the following: Given $ƒ(x) = e^x$, verify that $\lim_{h\to 0}\frac{e^{x+h} – e^x}{h} = e^x$. Given $ƒ(x) = e^x$, verify that 
$$\lim_{h\to 0}\frac{e^{x+h} – e^x}{h} = e^x$$  and explain how this illustrates that $f'(x) = \ln e \cdot f(x) = f(x)$.
 A: A proper answer to this question is fully dependent on the proper definition of symbol $e^{x}$. And further a proper definition of $e^{x}$ can not be given by just defining $e$.
One definition of $e^{x}$ is given by $$e^{x} = \lim_{n \to \infty}\left(1 + \frac{x}{n}\right)^{n}\tag{1}$$ Although it is normally self-evident because of the symbol $n$, I want to emphasize that the $n$ in the above definition is a positive integer. Based on this definition it can be proven that $$e^{x + y} = e^{x}\cdot e^{y}\tag{2}$$ and $$\lim_{x \to 0}\frac{e^{x} - 1}{x} = 1\tag{3}$$ Both the equations $(2)$ and $(3)$ are proved in this answer based on definition $(1)$. Using these equations $(2), (3)$ it is easy to prove that $$\frac{d}{dx}\,e^{x} = e^{x}\tag{4}$$ Second part of the question requires us to define the symbol $\ln$ (or $\log$ properly). One such definition is that if $e^{y} = x$ then $y = \ln x$ and then $\ln e = 1$ and this solves the second part of your question.
A: Recall that all we need to show is that $\lim_{h\to 0} \frac{e^h - 1}{h} = 1$.  The only things we need in this case are the definition of $e$ given by $e^h = \lim_{n\to \infty}\left ( 1 + \frac{h}{n} \right )^n$, and the binomial theorem.  Observe that 
\begin{eqnarray*}
\lim_{h\to 0} \frac{e^h -1}{h} & = & \lim_{h\to 0} \lim_{n\to \infty} \frac{1}{h} \left ( 1 + \frac{h}{n} \right )^n - \frac{1}{h} \\
& = & \lim_{h\to 0}\lim_{n\to \infty} \frac{1}{h} \sum_{k=0}^n \binom{n}{k}\left ( \frac{h}{n}\right )^k - \frac{1}{h}.
\end{eqnarray*}
If we factor out $\frac{1}{h}$ from both terms, we see that we can simply start the summation from the $k=1$ term, hence we have 
\begin{eqnarray*}
\lim_{h\to 0} \frac{e^h-1}{h} & = & \lim_{h\to 0} \lim_{n\to \infty} \frac{1}{h} \sum_{k=1}^n \binom{n}{k} \left ( \frac{h}{n} \right )^k \\
& = & \lim_{h\to0} \lim_{n\to \infty}\frac{1}{h} \left (n \frac{h}{n} + \frac{n(n-1)}{2}\frac{h^2}{n^2} + \frac{n(n-1)(n-2)}{3!}\frac{h^3}{n^3} + \ldots + \frac{h^n}{n^n} \right ) \\
& = & \lim_{h\to 0} \lim_{n\to\infty} \left( 1 + \binom{n}{2}\frac{h}{n} + \binom{n}{3} \frac{h^2}{n^3} + \ldots + \frac{h^{n-1}}{n^n} \right ).
\end{eqnarray*}
No matter what the form of the series is in the limit we'll have that there are terms in various powers of $h$, which in the limit as $h\to 0$ will all vanish.  The only term that remains is the value of $1$ at the beginning of the summation.
A: Using rules of exponentials,
$$\lim_{h \rightarrow 0} \frac{e^{x+h}-e^x}{h} = \lim_{h \rightarrow 0} \frac{e^xe^h-e^x}{h}$$
$$\lim_{h \rightarrow 0} \frac{e^{x+h}-e^x}{h} = e^x \lim_{h \rightarrow 0} \frac{e^h-1}{h}$$
This has a limit of $1$ (although I'm not sure how to show this without using L'Hospital's rule. Therefore,
$$\lim_{h \rightarrow 0} \frac{e^{x+h}-e^x}{h} = e^x$$
