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I'm looking for a straight forward proof using the definition of a derivative applied to the exponential function and substitution of one of the limit definitions of $e$, starting with

$e = \lim_{h\to \infty}\left({1+\dfrac{1}{h}}\right)^h$ or $e=\sum_{h=0}^{\infty}{\dfrac{1}{h!}}$


$\dfrac{d}{dx}\left( e^x \right) = \lim_{h\to 0}\left({\dfrac{e^{x+h}-e^{x}}{h}}\right)$

I found a proof I sort of liked here (which is sort of along the lines of a proof I'd like to use):


My only problem is that he combines the dummy variable, $h$, for the limit definition of $e$ and the dummy variable, $h$, used for the derivative. To me, it seems like it's not quite valid to do such a thing because it assumes both values are equal. Can anyone provide a better proof or justification for why the dummy variables can be combined?


I guess I'd also like to have a proof of why:

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

using one of the limit definitions of $e$ shown above.

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For the last limit, you may use the definition of $e$ you wrote first, i.e. $e^h=\sum_{k\geq 0} \frac{h^k}{k!}$. –  Martingalo Feb 10 '14 at 19:36
@Martingalo I'd prefer not to go that route. –  user1346994 Feb 10 '14 at 19:38
The first definition of $\;e\;$ requires $\;h\to \infty\;$ , not to zero. –  DonAntonio Feb 10 '14 at 19:38
What is your definition of the function $x \mapsto e^x$? –  Svinepels Feb 10 '14 at 19:39
The comment by @Svinepels is to the point: Defining $e^x$ is the hard part. The common approaches to doing so have the derivative formula as an easy corollary. –  Harald Hanche-Olsen Feb 10 '14 at 19:41

2 Answers 2

$$\frac{e^h-1}h=\frac1h\left(h+\frac{h^2}{2!}+\frac{h^3}{3!}+\ldots\right)=1+\frac h{2!}+\frac{h^2}{3!}+\ldots\xrightarrow[h\to 0]{}1$$

Of course, some power series theory must be known to fully justify the above. And now all it's easy:

$$\lim_{h\to 0}\frac{e^{x+h}-e^x}h=\lim_{h\to 0}\,e^x\frac{e^h-1}h=e^x\cdot 1=e^x$$

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Ok, I saw now what you mean but it is not a problem: Define $$e=\lim_{t\to 0} (1+t)^{1/t}$$ then $$e^h = \lim_{t\to 0} \left( (1+t)^{1/t} \right)^h$$ (I could do it because the function $x^h$ is continuous, so you can take the lim out)


$$\lim_{h\to 0} \frac{e^h -1}{h} = \lim_{h\to 0} \frac{\lim_{t\to 0} \left( (1+t)^{1/t} \right)^h -1}{h} = \lim_{t\to 0}\lim_{h\to 0} \frac{(1+t)^{h/t} -1}{h} $$ again by continuity.

Now make the change of variables: $y=h/t$, then $y\to 1$ and $h\to 0$. So

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

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