2 of 3 Minor Edit. Fixed Row Alignment.

Hint: For $$h\neq0$$, we have $$~\displaystyle\int_1^xt^{~h-1}~dt=\bigg[\frac{t^h}h\bigg]_1^x=\frac{x^h-1}h.~$$ For $$h=0$$, we have $$~\displaystyle\int_1^x\frac{dt}t=$$

$$=\displaystyle\lim_{h\to0}\frac{x^h-1}h.~$$ At the same time, $$~\Big(a^x\Big)'=\displaystyle\lim_{h\to0}\dfrac{a^{x+h}-a^x}h=a^x~\lim_{h\to0}\dfrac{a^h-1}h.~$$ Do you notice

anything “suspicious” ? ;-) Let e be the number for which this limit is $$1$$, and let $$\ln=\log_{~\large e}.$$ Then it

follows that $$~\Big(a^x\Big)'=\Big(e^{x\ln a}\Big)'=(x\ln a)'\cdot e^{x\ln a}\cdot1=a^x\ln a.~$$ Thus, $$~\displaystyle\int_1^x\frac{dt}t=\ln x,$$ which

for $$x=e,~$$ yields $$~\displaystyle\int_1^{\large e}\frac{dt}t=1.~$$ Now, let us inspect the numbers $$u(h)$$ for which $$~\displaystyle\int_1^{u(h)}t^{h-1}~dt$$

$$=1.~$$ Integrating, we have $$\dfrac{u^h-1}h=1\iff u=\sqrt[^h]{1+h}.~$$ Letting $$h\to0,~$$ we have $$e=u(0)$$

$$=\displaystyle\lim_{h\to0}(1+h)^{^\tfrac1h}=\lim_{n\to\infty}\bigg(1+\dfrac1n\bigg)^n$$. Hope this helps.

 answered Aug 20 '14 at 21:25 Lucian 41.7k11 gold badge6060 silver badges131131 bronze badges