Prove the definition of e

**wopashui** · February 5th 2010, 04:28 PM

Prove that
$lim_{h-->0}$ $(1+h)^{1/h}$ = 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

**Prove It** · February 5th 2010, 04:39 PM

Originally Posted by wopashui

Prove that
$lim_{h-->0}$ $(1+h)^{1/h}$ = 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 $e^x$ to be the function that is its own derivative. We want to try to find what value $e$ is.

So let $f(x) = e^x$

$f'(x) = \lim_{h \to 0}\frac{f(x + h) - f(x)}{h}$ .

So $f'(x) = \lim_{h \to 0}\frac{e^{x + h} - e^x}{h}$

$= \lim_{h \to 0}\frac{e^xe^h - e^x}{h}$

$= \lim_{h \to 0}\frac{e^x(e^h - 1)}{h}$

$= e^x \lim_{h \to 0}\frac{e^h - 1}{h}$ .

So for $f(x)$ to be equal to $f'(x)$ , we would require

$\lim_{h \to 0}\frac{e^h - 1}{h} = 1$

$\lim_{h \to 0}(e^h - 1) = \lim_{h \to 0}h$

$\lim_{h \to 0}e^h - \lim_{h \to 0}1 = \lim_{h \to 0}h$

$\lim_{h \to 0}e^h = \lim_{h \to 0}1 + \lim_{h \to 0}h$

$\lim_{h \to 0}e^h = \lim_{h \to 0}(1 + h)$

$\lim_{h \to 0}(e^h)^{\frac{1}{h}} = \lim_{h \to 0}(1 + h)^{\frac{1}{h}}$

$\lim_{h \to 0}e = \lim_{h \to 0}(1 + h)^{\frac{1}{h}}$

$e = \lim_{h \to 0}(1 + h)^{\frac{1}{h}}$ .

**Plato** · February 5th 2010, 04:58 PM

Originally Posted by wopashui

Prove that
$lim_{h-->0}$ $(1+h)^{1/h}$ = 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

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 $e$ is defined as the number such that $\int_1^e {\frac{{dx}}{x}} = 1$

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