6
$\begingroup$

A common definition of $e$ is given as $e = \lim_{n\rightarrow\infty}\left(1+\frac{1}{n}\right)^{n}$ which can be proven to be equivalent to $e=\lim_{h\rightarrow 0}\ \left(1+h\right)^{\frac{1}{h}}$ The most practical use of $e$ in elementary calculus is however given as $1=\lim_{h\rightarrow 0}\frac{e^h - 1}{h}$ which is used as a statement the slope of $e^x$ at $x=0$ is $1$ allowing one to prove that $\frac{d}{dx}e^x=e^x$. It appears trivial to prove that the two limits given above are equivalent, but I cannot seem to make any progress without making some illegal limit operations. I suspect the problem is deeper than it appears (I suspect the trouble is that although we have defined $e$, we have not actually said anything about what $e^x$ is). How does one rigorously proceed from the given definition of $e$ to the slope limit?

  • 0
    I would go with using the binomial expansion, you see...2011-12-03

2 Answers 2

1

It is easy to go from the first limit to the second and back by using the substitution

$z=\ln(1+h) \,,$

Note that this is equivalent to $h= e^{z}-1$.

and the continuity of the $\ln()$.

$e=\lim_{h\rightarrow 0}\ \left(1+h\right)^{\frac{1}{h}} \Leftrightarrow $ $1=\lim_{h\rightarrow 0}\ {\frac{\ln \left(1+h\right)}{h}} \Leftrightarrow$ $1=\lim_{z\rightarrow 0}\ {\frac{z}{e^z-1}} $

Writing down a formal proof is easy now, you just have to be carefull filling it the details. You need to use the continuity of the logarithm function, the fact that $z=\ln(1+h)$ is a bijection from a neighborhood of zero to a neighborhood of zero; and most importantly that for each of the two implications $e$ and hence $\ln$ are defined in a different way. Thus, you cannot really work with if and only if, because then $\ln$ makes no sense. But you can do each implication separately easely...

1

I'm not sure if you have multiple questions here but I'll try to answer all of them. For deriving: $1=\lim_{h\rightarrow 0}\frac{e^h - 1}{h}$

We start with the definition of a derivative:

f'(x)=\lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}

If we let $f(x)=a^{x}$, where $a \in \mathbb{R}$ then,

f'(x) = \lim_{h \rightarrow 0} \frac{a^{x+h} - a^{x}}{h} f'(x) =\lim_{h \rightarrow 0} \frac{a^{x}(a^{h} - 1)}{h} f'(x) =a^{x} \left( \lim_{h \rightarrow 0} \frac{a^{h} - 1}{h} \right) If we let $a=e$, since $e \in \mathbb{R}$ we have that f'(x) = e^{x} \left( \lim_{h \rightarrow 0} \frac{e^{h} -1}{h} \right) Using l'hopital's rule and that $e^{h}$ is continuous at $h=0$ we can say that, f'(x)= e^{x} \ln(e) So, if we evaluate it at $x=0$ we have, f'(0)=e^{0}=1 So that is how you obtain it one way.

Another way would go as follows: Let $f(x)=\ln(x)$, then f'(x) = \frac{1}{x}, so f'(1)=1 and then we say, $1= \lim_{h \rightarrow 0} \frac{f(1+h) - f(1)}{h}$ Because I like x's $1 = \lim_{x \rightarrow 0} \frac{f(1+x) - f(1)}{x}$ $1 = \lim_{x \rightarrow 0} \frac{\ln(1+x) - \ln(1)}{x}$ $1 = \lim_{x \rightarrow 0} \frac{1}{x} \ln(1+x)$ $1 = \lim_{x \rightarrow 0} \ln(1+x)^{\frac{1}{x}}$ Raise both sides by $e$ and we get, $e^{1} = e^\left({\lim_{x \rightarrow 0} \ln(1+x)^{\frac{1}{x}}}\right)$ $e = \lim_{x \rightarrow 0} \text{ }e^\left({\ln(1+x)^{\frac{1}{x}}}\right)$ $e = \lim_{x \rightarrow 0} (1+x)^{\frac{1}{x}}$ And there you have your second formulation. For the first one, just let $x = \frac{1}{n}$ and you receive, $e = \lim_{n \rightarrow \infty} \left(1 + \frac{1}{n} \right)^{n}$

Hopefully that helps! Let me know if I made a mistake somewhere.

  • 0
    Where does it circular? You start with the fact that $\frac{d}{dx} \ln(x) = \frac{1}{x}$ and from there you derive the definition of e.2011-12-19