In this YouTube video it is said that $ e $ naturally arises as a number that allows us to take the derivatives of functions like $ a^x $. So $ e $ is defined as a number for which: $$(e^x)'=e^x\lim_{h\rightarrow 0}\frac {e^h-1}{h}=e^x$$

So essentially the limit should be $1 $ for this to happen. How would you prove that there is indeed such a real number, and find its value?

P.S.Please, don't say that there are other ways to define $ e $ - I know that. It's just this particular definition that seems natural and interesting to me.

  • $\begingroup$ Since you're using the "arithmetic" definition of $e^x$, to calculate that derivative you'll first need to prove it converges, covered in this question. $\endgroup$ – Jack M May 6 '14 at 19:19

If you had a definition of $a^b$ (which most books base on having a definition of $\ln$ and $e$, so that $a^b := e^{b \ln a}$), you might say to yourself

  1. $a^b$ depends continuously on $a$ and $b$ as long as $a$ and $b$ are positive.

  2. When I compute $\lim_{h \to 0} \frac{2^h - 1}{h}$, I get a number less than 1.

  3. When I compute $\lim_{h \to 0} \frac{4^h - 1}{h}$, I get a number greater than 1.

Therefore by the intermediate value theorem, there should be a number $c$ with $$ \lim_{h \to 0} \frac{c^h - 1}{h} = 1. $$

This depends, of course, on the well-definedness and continuity of the function

$$ F(b) =\lim_{h \to 0} \frac{b^h - 1}{h} $$

which I cannot prove without either a very messy epsilon-delta argument, or relying on alternative definitions of $e$. But this is at least the gist of how you might go about things. I'd love to see your definition of $a^b$ for irrational $a$ and $b$, though, so that I could turn it into a real proof.

  • $\begingroup$ John's first displayed equation implies that the desired value of $c$ is $\lim_{h \rightarrow 0} (1+h)^\frac{1}{h}$, which is indeed one way to define Euler's $e$. $\endgroup$ – Austin Mohr May 6 '14 at 19:24
  • $\begingroup$ My definition of $ a^b $ is as given in this question. $\endgroup$ – user132181 May 6 '14 at 19:26
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    $\begingroup$ Yeah...with that definition, I need to know how you defined "log" (and how you prove the existence of $\log(x)$ for every positive $x$). I have a feeling that I would end up in circular-definition hell, so I won't pursue this. It's possible that someone else could do so profitably. $\endgroup$ – John Hughes May 7 '14 at 18:28

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