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
