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I tried to search for a few minutes but I didn't find this question so I hope it's not a duplicate.

So I want to show that $(e^x)' = e^x$. To do that, I must proof that the limit:
$$\lim_{h\to0}\frac{f(x+h) - f(x)}{h} = \lim_{h\to0}\frac{e^{x+h} - e^x}{h}$$ exists and equals to $e^x$.

So I have:
$$\frac{e^{x+h} - e^x}{h} = \frac{e^x \cdot e^h - e^x}{h} = e^x \bigg(\frac{e^h - 1}{h}\bigg) \\ e^h - 1 = z \implies e^h = 1+z >0 \implies h = \ln(1+z)$$ Because of the continuity of the $\ln$ and $e^x$ functions, we have:
$$h\to 0 \iff e^h \to 1 \implies z\to0 \\ z > 0 \implies \frac{1}{z} \to +\infty \\ z < 0 \implies \frac{1}{z} \to -\infty \\ \frac{e^h - 1}{h} = \frac{z}{\ln(1+z)} = \frac{1}{\frac{1}{z}\ln(1+z)} = \bigg[ y = \frac{1}{z} \bigg] = \frac{1}{y\ln(1+\frac{1}{y})} = \frac{1}{\ln(1+\frac{1}{y})^y} \\ h \to 0 \implies |y| \to +\infty \implies \bigg(1+\frac{1}{y}\bigg)^y \to e \implies \\ \lim_{h\to0}\frac{e^h - 1}{h} = \frac{1}{\ln e} = 1 \implies (e^x)' = e^x \cdot 1 = e^x$$

But how do I prove that $\lim_{\pm\infty}(1 + \frac{1}{x})^x = e$ ?

We defined $e$ like this: $$\lim_{n\to\infty} \bigg(1 + \frac{1}{n}\bigg)^n = e \space \,, \space n\in\mathbb{N}$$

I thought of the sandwich theorem:

$\forall x \geq 1 \,, x\in\mathbb{R} \,, \lfloor x \rfloor \leq x < \lfloor x \rfloor + 1$ and $\lfloor x \rfloor , \lfloor x \rfloor + 1 \in \mathbb{N} $ and if I denote $\mathbb{N} \ni n = \lfloor x \rfloor$ I have: $$ \bigg(1 + \frac{1}{n+1}\bigg)^n \leq \bigg(1 + \frac{1}{x}\bigg)^x \leq \bigg(1 + \frac{1}{n}\bigg)^{n+1} \\ \lim_{n\to+\infty}\bigg(1 + \frac{1}{n+1}\bigg)^n = \frac{\lim_{n\to+\infty}\bigg(1 + \frac{1}{n+1}\bigg)^{n+1}}{\lim_{n\to+\infty}\bigg(1 + \frac{1}{n+1}\bigg)} = \frac{e}{1+0} = e \\ \lim_{n\to+\infty}\bigg(1 + \frac{1}{n}\bigg)^{n+1} = \lim_{n\to+\infty}\bigg(1 + \frac{1}{n}\bigg)^n \cdot \lim_{n\to+\infty}\bigg(1 + \frac{1}{n}\bigg) = e \cdot (1 + 0) = e$$

and now I could apply the theorem but then again, how do I show that this: $\bigg(1 + \frac{1}{n+1}\bigg)^n \leq \bigg(1 + \frac{1}{x}\bigg)^x \leq \bigg(1 + \frac{1}{n}\bigg)^{n+1}$ is true? I mean, I can say that (left side): $$ n+1 > x \implies \frac{1}{n+1} < \frac{1}{x} \implies 1 + \frac{1}{n+1} < 1 + \frac{1}{x} \\ \forall y\in\mathbb{R_+} \space n \leq x \implies y^n \leq y^x \\ \implies \bigg(1 + \frac{1}{n+1}\bigg)^n \leq \bigg(1 + \frac{1}{x}\bigg)^x$$ and by analogy the right side but is this a proof in terms of limits?

and that's just $+\infty$ for $x \geq 1$, what do I do for $-\infty$ ?

For $x\in\langle 0, 1\rangle$, a substitution $y = \frac{1}{x}$ gives me: $$\lim_{y\to 0}\bigg(1 + y\bigg)^\frac{1}{y}$$

I'll stop here to see if I'm on the right track, any hints, suggestions, edits, comments and answers are welcome!

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  • $\begingroup$ Here is a proof for $(e^x)'=e^x$. $\endgroup$ – Hakim May 20 '14 at 17:37
  • $\begingroup$ Proving that e is the limit of that expression seems to involve LHR which I guess is not allowed. Perhaps epsilon-N? $\endgroup$ – BCLC May 20 '14 at 17:37
  • $\begingroup$ @Shirohige math.stackexchange.com/a/631980/140308 $\endgroup$ – BCLC May 20 '14 at 17:46
  • $\begingroup$ I like to get around all of this by simply defining $e^x$ to be the solution of the ode $y'(x)=y(x)$ such that $y(0)=1$. $\endgroup$ – JP McCarthy May 20 '14 at 17:48
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    $\begingroup$ @BCLC Don't we have to prove that both functions are differentiable before we can apply the chain rule on them? $\endgroup$ – AltairAC May 20 '14 at 17:54
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We will use the basic derivative definition: $$\lim_{h\to0}\frac{f(x+h) - f(x)}{h} = \lim_{h\to0}\frac{e^{x+h} - e^x}{h}.$$ After some algebraic manipulations you will find that it is equal to: $$e^x\cdot\color{blue}{\lim_{h\to0}\frac{e^h-1}h}.$$ Here's a proof that $\color{blue}{\rm this\, term}$ is equal to one, hence $(e^x)'=e^x$, taken from proofWiki:

$$\eqalign{ \frac{e^ h - 1} h&= \frac{\lim\limits_{n \to \infty} \left({1 + \dfrac h n}\right)^n - 1} h \\ &= \frac{\displaystyle \lim_{n \to \infty} \sum_{k \mathop = 0}^n {n \choose k} \left({\frac h n}\right)^k - 1} h\\ &= \lim_{n \to \infty} \frac{\displaystyle \sum_{k \mathop = 0}^n {n \choose k} \left({\frac h n}\right)^k - 1} h\\ &= \lim_{n \to \infty} \left({ {n \choose 0} 1 - 1 + {n \choose 1} \left({\frac h n}\right)\frac 1 h + \sum_{k \mathop = 2}^n {n \choose k} \left({\frac h n}\right)^k \frac 1 h }\right)\\ &=\lim_{n \to \infty}1 + \lim_{n \to \infty} \sum_{k \mathop = 2}^n {n \choose k} \frac {h^{k-1} }{n^k}\\ &=1 + h \lim_{n \to \infty} \sum_{k \mathop = 2}^n {n \choose k} \frac{h^{k-2} } {n^k}.\\ }$$ The right summand converges to zero as $h\to0$, and so: $$\lim_{h\to0}\frac{e^h-1}h=1.$$

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    $\begingroup$ Wow, this is an amazingly efficient and simple answer, thank you! But if you don't mind I would like to wait some time to see if my approach can be applied (I spent a lot of time writing this question and I would like to understand "my proof" if it's valid or invalid and what needs to be added (if the approach is valid of course)...) $\endgroup$ – AltairAC May 20 '14 at 17:57
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    $\begingroup$ This proof depends on $\lim_{n\to\infty}\sum_{k=2}^n{n\choose k}\frac{h^{k-2}}{n^k}$ being a bounded function of $h$ (otherwise the summand need not converge to zero as $h\to 0$). $\endgroup$ – vadim123 May 20 '14 at 18:21

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