Procedure to Get Derivative of exp(x) Respect to x

[trfrm]

Horas.

Excuse me ... .

The definition of derivative a function $f(x)$ respect to $x$ is

$\frac{d}{dx}f(x):=\displaystyle\lim_{\Delta{x}\rightarrow0}\frac{f(x+\Delta{x})-f(x)}{\Delta{x}}$ ... .

Consider $f(x)=\exp(x)$ ... then

$\frac{d}{dx}\exp(x)=\exp(x)\displaystyle\lim_{\Delta{x}\rightarrow0}\frac{\exp(\Delta{x})-1}{\Delta{x}}$ ... .

In real calculus, we have known that $\frac{d}{dx}\exp{x}=\exp{x}$ ... .

So, we have to say that

$\displaystyle\lim_{\Delta{x}\rightarrow0}\frac{\exp(\Delta{x})-1}{\Delta{x}}=1$ ... (*) ... .

The problem is how to prove the last equality (*) ... .

Maybe, we can prove the last equality by L'Hopital Theorem ... .

But, the L' Hopital Theorem uses the concept of derivative ... .

Whereas, the concept of derivative uses the concept of limit ... .

How can the last equality (*) be proven without L'Hopital Theorem ... ?

Is the last equation (*) an assumption ... ?

Thank you very much ... .

Gloria in excelsis Deo.

[trfrm]

Salam damai Kristus.

Excuse me ... .

I have known that the definition of Euler number is $\textrm{e}:=\displaystyle\lim_{\epsilon\rightarrow0}(1+\epsilon)^{1/\epsilon}$ ... .

If we use the power series procedure, we must use concept of derivative ... because the in Maclaurin series exist derivative expression, that is

$f(x)=\displaystyle\sum_{j=0}^{\infty}\frac{1}{j!}\left(\frac{d\,f(x)}{dx}\right)_{x=0}x^j$ ... .

Using Maclaurin series, we have to seek derivative of $\textrm{e}^x$ ... , whereas we will prove that derivative of $\textrm{e}^x$ is $\textrm{e}^x$ ... .

I'm sorry if I do mistakes ... . Thank you for the attention ... .

Sanctus, Sanctus, Dominus Deus Sabaoth.