Is required to prove the derivative of [latex]e^x[/latex] by definition, so

 

[latex]\displaystyle\frac{d}{dx}(e^x)=\lim_{\Delta x \to 0}{\displaystyle\frac{e^{\Delta x +x}-e^x}{\Delta x}}=e^x\lim_{\Delta x \to 0}{\displaystyle\frac{e^{\Delta x}-1}{\Delta x}}[/latex]

 

I can also express the exponential function as [latex]e^{\Delta x}=\sum_{n=0}^{\infty}{\displaystyle\frac{{\Delta x}^n}{n!}}[/latex]

 

[latex]e^{\Delta x}-1=\sum_{n=1}^{\infty}{\displaystyle\frac{{(\Delta x)}^n}{n!}}=\sum_{n=0}^{\infty}{\displaystyle\frac{{(\Delta x)}^{n+1}}{(n+1)!}}[/latex]

 

How i can continue?

After division by Δx, the first term is 1, while the remaining terms -> 0.

Is required to prove the derivative of [latex]e^x[/latex] by definition, so

 

[latex]\displaystyle\frac{d}{dx}(e^x)=\lim_{\Delta x \to 0}{\displaystyle\frac{e^{\Delta x +x}-e^x}{\Delta x}}=e^x\lim_{\Delta x \to 0}{\displaystyle\frac{e^{\Delta x}-1}{\Delta x}}[/latex]

 

I can also express the exponential function as [latex]e^{\Delta x}=\sum_{n=0}^{\infty}{\displaystyle\frac{{\Delta x}^n}{n!}}[/latex]

 

[latex]e^{\Delta x}-1=\sum_{n=1}^{\infty}{\displaystyle\frac{{(\Delta x)}^n}{n!}}=\sum_{n=0}^{\infty}{\displaystyle\frac{{(\Delta x)}^{n+1}}{(n+1)!}}[/latex]

 

How i can continue?

 

 

We have that: [math]lim_{x\to 0}\frac{e^x-1}{x} = 1[/math]

 

 

if 0<x<1,then we have:[math]1\leq\frac{e^x-1}{x}\leq\frac{1}{1-x}[/math]....................................................................1

 

 

if -1<x<0 ,then we have: [math]\frac{1}{1-x}\leq\frac{e^x-1}{x}\leq1[/math]..................................................................2

 

And we observe that : [math]\lim_{x\to 0}\frac{1}{1-x} = 1[/math] ,hence [math]lim_{x\to 0}\frac{e^x-1}{x} = 1[/math]

 

The proof of the inequalities (1) and (2) is based on the fact that : [math] \forall x(x\in R\Longrightarrow e^x\geq 1+x)[/math]