Looking to prove $ f(x) = e^x =f'(x) $ using the limit definition of the derivative. So far I have gotten to here:
$e^x * \lim\limits_{h \to 0} \frac{e^h-1}{h}$
Any hints on what I can do next to get that entire limit to $1$?
I can't use l'hopital's rule, otherwise that'd defeat the purpose.
$$\large e=\lim_{\Delta x \to 0 }(1+\Delta x)^{\frac{1}{\Delta x}}\space (definition)\\ e^x=\lim_{\Delta x \to 0 }(1+\Delta x)^{\frac{x}{\Delta x}}\\ f'(x)=\lim_{\Delta x \to 0 }\frac{f(x+\Delta x)-f(x)}{\Delta x }=\\ \large \lim_{\Delta x \to 0 }\frac{(1+\Delta x)^{\frac{x+\Delta x}{\Delta x}}-(1+\Delta x)^{\frac{x}{\Delta x}}}{\Delta x }=\\ \large \lim_{\Delta x \to 0 }\frac{(1+\Delta x)^{\frac{x}{\Delta x}}((1+\Delta x)^{\frac{\Delta x}{\Delta x}}-1)}{\Delta x }=\\ \large \lim_{\Delta x \to 0 }\frac{(1+\Delta x)^{\frac{x}{\Delta x}}((1+\Delta x)^1-1)}{\Delta x }=\\ \large \lim_{\Delta x \to 0 }\frac{(1+\Delta x)^{\frac{x}{\Delta x}}(\Delta x)}{\Delta x }=\\ \large \lim_{\Delta x \to 0 }(1+\Delta x)^{\frac{x}{\Delta x}}=\\e^x$$