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I couldn't find this exact question, apologize if it's a duplicate.

I would like to show based only on the above definition of e, that this equality for $e^x$ holds, without going through showing it's a bijective function, has an inverse function ln, finding it's derivative and finding it's taylor series. I've tried to play with it, but except for x=-1 (with a proof that doesn't seem to be possible to generalize), I had no progress.

Is their a straightforward proof for that? Or does one has to go through the whole process mentioned above?

Merkh
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galra
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  • What is your definition of $e^x$, if it (evidently) isn't what you're trying to prove? – Brian Tung Oct 17 '16 at 02:20
  • Or are you only trying to show it for integer values of $x$? – Brian Tung Oct 17 '16 at 02:21
  • It's possible to do it without the method you describe above, say by showing that if $x \in \mathbb{R}$, then you can prove that $\sum_{n=0}^{\infty} x^n/n!$ converges to $\lim (1+x/n)^n$. This is done or sketched, if I recall correctly, in Baby Rudin in the sequences/series chapter, for example. – Geoff Oct 17 '16 at 02:22

2 Answers2

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Extending the Limit Over the Integers to the Limit Over the Reals

For $\alpha\ge1$, $$ \underbrace{\left(1+\frac1{\lfloor\alpha\rfloor+1}\right)^{\large\lfloor\alpha\rfloor}}_{\color{#C00000}{\left(1+\frac{\large1}{\lfloor\alpha\rfloor+1}\right)^{-1}}\color{#00A000}{\left(1+\frac{\large1}{\lfloor\alpha\rfloor+1}\right)^{\large\lfloor\alpha\rfloor+1}}} \le\left(1+\frac1\alpha\vphantom{\frac1{\lfloor\alpha\rfloor}}\right)^{\large\alpha} \le\underbrace{\left(1+\frac1{\lfloor\alpha\rfloor}\right)^{\large\lfloor\alpha\rfloor+1}}_{\color{#C00000}{\left(1+\frac{\large1}{\lfloor\alpha\rfloor}\right)}\color{#00A000}{\left(1+\frac{\large1}{\lfloor\alpha\rfloor}\right)^{\lfloor\alpha\rfloor}}}\tag{1} $$ where the red terms tend to $1$ and the green terms tend to $e$ by the standard limit $$ \lim_{n\to\infty}\left(1+\frac1n\right)^n=e\tag{2} $$ where $n\in\mathbb{Z}$. Therefore, $(1)$ and the Squeeze Theorem says that $$ \lim_{\alpha\to\infty}\left(1+\frac1\alpha\right)^{\large\alpha}=e\tag{3} $$ where $\alpha\in\mathbb{R}$. Furthermore, $(3)$ implies $$ \begin{align} \lim_{\alpha\to\infty}\left(1-\frac1\alpha\right)^{\large-\alpha} &=\lim_{\alpha\to\infty}\left(1+\frac1{\alpha-1}\right)^{\large\alpha}\\ &=\lim_{\alpha\to\infty}\left(1+\frac1{\alpha-1}\right)^{\large\alpha-1}\left(1+\frac1{\alpha-1}\right)\\[9pt] &=e\cdot1\tag{4} \end{align} $$ Thus, $(3)$ and $(4)$ give $$ \lim_{|\alpha|\to\infty}\left(1+\frac1\alpha\right)^{\large\alpha}=e\tag{5} $$

Applying $\boldsymbol{(5)}$ to the Question $$ \begin{align} \lim_{n\to\infty}\left(1+\frac xn\right)^n &=\lim_{n\to\infty}\left(1+\frac xn\right)^{\large\frac nx\cdot x}\\[9pt] &=e^x\tag{6} \end{align} $$ for all $x$.

robjohn
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Here, by substitution...

$$\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n = \lim_{n\to\infty}\left(1+\frac{1}{\frac{n}{x}}\right)^n = \lim_{m\to\infty}\left(1+\frac{1}{m}\right)^{xm} = \lim_{m\to\infty}\left(\left(1+\frac{1}{m}\right)^m\right)^x = e^x$$

This will only work for positive x, but if you've already got it for -1, you should be able to take it from there.

user361424
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  • Oh, no! The OP only knows the limit of the sequence, you're computing the limit over the reals, which is a different matter. – egreg Oct 17 '16 at 11:17