# An alternative second half?

Going back to the previous blog and looking for an alternative second half…

We have $\frac{d}{dx}e^x =^{\lim_\delta \to 0}_{\lim_ n \to 0} (1 + n)^{\frac{x}{n}}\frac{(1+n)^{\frac{\delta}{n}}-1}{\delta}$

For this to be true $\frac{(1+n)^{\frac{\delta}{n}}-1}{\delta}=1$

So $\delta = (1+n)^{\frac{\delta}{n}}-1$

$\delta + 1 = (1+n)^{\frac{\delta}{n}}$

Raise both sides to the $n$th power:

$(\delta + 1)^n = (1+n)^\delta$

At $\lim_{\delta \to 0}$ and $\lim_{n \to 0}$ it can be seen that both sides equal 1.

Feels like quite a weak proof to me. But I am not in a position to claim expertise. Thoughts, anyone?