Prove that
$\displaystyle lim_{h-->0}$ $\displaystyle (1+h)^{1/h} $= e
given that: If g is continuous at c and f is continuous a g(c), then the composition f o g is continuous at c
By definition, we want $\displaystyle e^x$ to be the function that is its own derivative. We want to try to find what value $\displaystyle e$ is.
So let $\displaystyle f(x) = e^x$
$\displaystyle f'(x) = \lim_{h \to 0}\frac{f(x + h) - f(x)}{h}$.
So $\displaystyle f'(x) = \lim_{h \to 0}\frac{e^{x + h} - e^x}{h}$
$\displaystyle = \lim_{h \to 0}\frac{e^xe^h - e^x}{h}$
$\displaystyle = \lim_{h \to 0}\frac{e^x(e^h - 1)}{h}$
$\displaystyle = e^x \lim_{h \to 0}\frac{e^h - 1}{h}$.
So for $\displaystyle f(x)$ to be equal to $\displaystyle f'(x)$, we would require
$\displaystyle \lim_{h \to 0}\frac{e^h - 1}{h} = 1$
$\displaystyle \lim_{h \to 0}(e^h - 1) = \lim_{h \to 0}h$
$\displaystyle \lim_{h \to 0}e^h - \lim_{h \to 0}1 = \lim_{h \to 0}h$
$\displaystyle \lim_{h \to 0}e^h = \lim_{h \to 0}1 + \lim_{h \to 0}h$
$\displaystyle \lim_{h \to 0}e^h = \lim_{h \to 0}(1 + h)$
$\displaystyle \lim_{h \to 0}(e^h)^{\frac{1}{h}} = \lim_{h \to 0}(1 + h)^{\frac{1}{h}}$
$\displaystyle \lim_{h \to 0}e = \lim_{h \to 0}(1 + h)^{\frac{1}{h}}$
$\displaystyle e = \lim_{h \to 0}(1 + h)^{\frac{1}{h}}$.
I find that definition rather bazaar. By that I mean beginning students do not have the tools to deal with that abstraction.
My all time favorite calculus textbook is the first edition of Gillman & McDowell, 1972.
It is normal is size( not the dictionary size), does integral as betweeness, and has this definition of e.
The number $\displaystyle e$ is defined as the number such that $\displaystyle \int_1^e {\frac{{dx}}{x}} = 1$