The function $e^x$ is one of most important function,
where $e$ is called Napier's constant.
\[ e=\lim_{n\rightarrow \infty}\left(1+\frac{1}{n}\right)^n=2.71828182845\cdots \]
or
\[ \frac{1}{e}=\lim_{n\rightarrow \infty}\left(1-\frac{1}{n}\right)^n=0.367879\cdots \]
The function $e^x$ is defined as follow.
\[ e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!} \]
$e^x$ has many important properties. one is
\[ (e^x)'=e^x, \quad or\quad \frac{d}{dx}e^x=e^x \]
[proof]
By definitions,
$(e^x)'=\frac{d}{dx}e^x=\lim_{h\rightarrow 0}\frac{e^{x+h}-e^x}{h}$
It is, as
$\lim_{h\rightarrow 0}\frac{e^{x+h}-e^x}{h}=e^x\lim_{h\rightarrow 0}\frac{e^{h}-1}{h}=e^x$ , then
\[ \lim_{h\rightarrow 0}\frac{e^{h}-1}{h}=1 \]
is required. At first, if $n=\frac{1}{h}$, then
\[ e=\lim_{h\rightarrow 0}\left(1+h\right)^{1/h} \]
Thus,
\[ \log e=\log \lim_{h\rightarrow 0}\left(1+h\right)^{1/h} =\lim_{h\rightarrow 0}\log \left(1+h\right)^{1/h} \]
\[ \lim_{h\rightarrow 0}\frac{\log (1+h)}{h}=1 \]
So, if we put $1+h=e^t$ , then $h=e^t-1$ . As $h\rightarrow 0$ , $t\rightarrow 0$ .
\[ \lim_{t\rightarrow 0}\frac{\log e^t}{e^t-1}=1 \]
\[ \lim_{t\rightarrow 0}\frac{t\log e}{e^t-1}=1 \]
\[ \lim_{t\rightarrow 0}\frac{t}{e^t-1}=1 \]
\[ \lim_{t\rightarrow 0}\frac{1}{\left(\frac{e^t-1}{t}\right)}=1 \]
Then, changing $t$ to $h$, we will get
\[ \lim_{h\rightarrow 0}\frac{e^{h}-1}{h}=1 \]
