Math - again

I have added MathJax to Drupal to write better math formulas.
It reminds me of the very first website I ever made back in 1995 with plain html, line by line. It's a whole lot of extra formatting, and no WYSIWYG here, but it works. To test out things I reproduce a simple high school math proof.

Let's take the derivative of $ \ln \left(x\right) $
Remember that the definition of a derivative is as follows:
$ \displaystyle\displaystyle\lim _{x\rightarrow a}\dfrac {f\left( x\right) -f\left( a\right) }{x-a}=f'\left( a\right) $
In that case we can write
$ \displaystyle\lim _{x\rightarrow a}\dfrac {\ln \left( x\right) -\ln \left( a\right) }{x-a} $

Because we know that
$ \log \left( a\right) -\log \left( b\right) =\log \left( \dfrac {a}{b}\right) $ and $ \log \left( a\right) ^{n}=n\log \left( a\right) $
we can write the equation as follows:
$ \displaystyle\lim _{x\rightarrow a}\dfrac {\ln \left( \dfrac {x}{a}\right) }{x-a} $
which is the same as
$ \displaystyle\lim _{x\rightarrow a}\dfrac {1}{x-a}.\ln \left( \dfrac {x}{a}\right) $
$ \displaystyle\lim _{x\rightarrow a}\ln \left( \dfrac {x}{a}\right) ^{\dfrac {1}{x-a}} $

If we substitute $ x-a=p $ which implies $ x=a+p $ we can continue a bit more civilized, only realizing that if
$ x\rightarrow a $ then $ p\rightarrow 0 $
So the equation becomes
$ \displaystyle\lim _{p\rightarrow 0}\ln \left( \dfrac {a+p}{a}\right) ^{\dfrac {1}{p}} $
and can be simplified to
$ \displaystyle\lim _{p\rightarrow 0}\ln \left( 1+\dfrac {p}{a}\right) ^{\dfrac {1}{p}} $

This time we are stuck with $ \dfrac {p}{a} $ which we solve by substitution again:
$ \dfrac {p}{a}=\dfrac {1}{n} $
which means
$ \dfrac {1}{p}=\dfrac {n}{a}=n\dfrac {1}{a} $

And if $ p\rightarrow 0 $ then $ n\rightarrow \infty $

This leads us to
$ \displaystyle\lim _{n\rightarrow \infty }\ln \left( 1+\dfrac {1}{n}\right) ^{n\cdot \dfrac {1}{a}} $
and of course
$ \displaystyle\lim _{n\rightarrow \infty }\dfrac {1}{a}\ln \left( 1+\dfrac {1}{n}\right) ^{n} $

Some might have forgotten it but this contains the definition of $ e $, namely
$ e=\displaystyle\lim _{n\rightarrow \infty } \left( 1+\dfrac {1}{n}\right) ^{n} $

And with that we can simply conclude that we end up with
$ \dfrac {1}{a}.\ln \left( e\right) =\dfrac {1}{a} $
$ f'\left(x\right) =\dfrac {1}{x} $

It seems to work. Back to the studio, then.