Why Is There a Contradiction in the Proof of the Derivative of lnx?

[coki2000]

The proof of derivative of lnx??

Hi all,
In proof of derivative of lnx,

\frac{d}{dx}(lnx)=\lim_{\Delta x\rightarrow 0}\frac{\ln(x+\Delta x)-lnx }{\Delta x}= \lim_{\Delta x\rightarrow 0}\frac{1}{\Delta x}\ln(1+\frac{\Delta x}{x})=\lim_{\Delta x\rightarrow 0}\frac{x}{\Delta x}\frac{1}{x}\ln(1+\frac{\Delta x}{x})=\lim_{\Delta x\rightarrow 0}\frac{1}{x}ln(1+\frac{\Delta x}{x})^{(\frac{x}{\Delta x})}

For \frac{x}{\Delta x}=n
\lim_{\Delta x\rightarrow 0}ln(1+\frac{\Delta x}{x})^{(\frac{x}{\Delta x})}\Rightarrow \lim_{n\rightarrow\infty}\ln (1+\frac{1}{n})^{n}

So
(lnx)'=\frac{1}{x}ln(\lim_{n\rightarrow\infty}(1+\frac{1}{n})^{n})=\frac{1}{x}

so we take the \lim_{n\rightarrow\infty}(1+\frac{1}{n})^{n}=e

But in proof of \lim_{n\rightarrow\infty}(1+\frac{1}{n})^{n} we take the derivative of lnx is \frac{1}{x}
(https://www.physicsforums.com/showthread.php?t=348071)

Please explain to me this contradiction.
Thanks.

 

[l'Hôpital]

Do you mean the circular logic problem? e is defined as the number that makes the base log such that (log x)' = 1/x.

Alternatively, you could do a proof by the binomial theorem, and take the limit, you'll see the definition for e in series form.