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Show that, $$e=\lim_{x\to \infty} \left(1+\frac{1}{x}\right)^x $$

Is the same number that satisfies, $$\lim_{h\to0} \frac{e^{h}-1}{h} = 1 \tag{*}$$

You don't have to do that if it's too cumbersome, but (*) is used to find the derivatives of real exponential functions. I know that definitions are not to be proven but I'm looking for some derivation or intution for the claim that (*) is a legitimate definition for e, and furthermore, a proof that limit in (*) exists.

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    You should have a look at this answer http://math.stackexchange.com/a/1491888/72031 and the linked blogpost http://paramanands.blogspot.com/2014/05/theories-of-exponential-and-logarithmic-functions-part-3.html where I deal with the limit of $(a^{x} -1)/x$ as $x \to 0$. – 2017-02-17
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    Awesome, thank you. – 2017-02-17
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    See Apostol book. – 2017-02-17

3 Answers 3

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For the forward implication, given that $e = \lim_{x \to \infty} (1 + 1/x)^x$, we have for $n \in \mathbb{N}$

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

It s straightforward to show using Bernoulli's inequality that $(1+1/n)^n$ is increasing and $(1 + 1/n)^{n+1}$ is decreasing.

Hence,

$$\left(1 + \frac{1}{n} \right)^n \leqslant e \leqslant \left(1 + \frac{1}{n} \right)^{n+1},$$

and

$$1 \leqslant \frac{e^{1/n}-1}{1/n} \leqslant n\left[\left(1 + \frac{1}{n} \right)\left(1 + \frac{1}{n} \right)^{1/n}- 1\right] = (n+1)\left(1 + \frac{1}{n} \right)^{1/n}- n .$$

Using Bernoulli's inequality we have

$$(n+1)\left(1 + \frac{1}{n} \right)^{1/n} - n \leqslant (n+1) \left(1 + \frac{1}{n}\frac{1}{n}\right) - n = 1 + \frac{1}{n} + \frac{1}{n^2}.$$

Thus,

$$1 \leqslant \frac{e^{1/n}-1}{1/n} \leqslant 1 + \frac{1}{n} + \frac{1}{n^2}.$$

Applying the squeeze theorem we see that

$$\tag{*}\lim_{n \to \infty} \frac{e^{1/n}-1}{1/n} = 1.$$

Continuity and monotonicity of the exponential function $x \mapsto e^x$ can be used to show that (*) implies

$$\lim_{h \to 0} \frac{e^{h}-1}{h} = 1.$$

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    The nice bits of monotone magic :-) – 2017-02-17
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    @SimplyBeautifulArt: Thanks. Obviously a lot going on here to fill in the details for general continuous limits, so the reference to Paramanand Singh's blog is probably the best completion to the answer. – 2017-02-18
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    Although I have written a good number of posts on these things, your derivation for limit of $n(e^{1/n}-1)$ is new for me. +1 – 2017-02-18
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    BTW we can also use the inequality $(1+(1/n))^{n}\leq e\leq (1-(1/n))^{-n}$. – 2017-02-18
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    @ParamanandSingh: Thank you. I have certainly picked up more than a few nice tricks from your answers over time. – 2017-02-18
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Consider a sequence $a_n$ that satisfies

$$\frac{(a_n)^{1/n}-1}{1/n}=1$$

It then follows that

$$a_n=\left(1+\frac1n\right)^n$$

And as $n\to\infty$...

(a bit of monotone magic may be in order)

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    You do need to add that "bit of monotone magic". – 2017-02-17
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$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

I like the following procedure because it shows how to choose the 'best' logaritm base $\ds{\mrm{a}}$: By 'best' I mean the simplest $\ds{\log}$-derivative expression.

\begin{align} \totald{\log_{\,\mrm{a}}\pars{x}}{x} & \stackrel{\mrm{def.}}{\equiv} \lim_{h \to 0}{\log_{\,\mrm{a}}\pars{x + h} - \log_{\,\mrm{a}}\pars{x} \over h} = \lim_{h \to 0}{1 \over h}\log_{\,\mrm{a}}\pars{1 + {h \over x}} \\[5mm] & = {1 \over x}\log_{\,\mrm{a}}\pars{\lim_{h \to 0}\bracks{1 + {h \over x}}^{x/h}} \end{align} The 'simplest $\ds{\,\mrm{a}}$-choice' is given by: $$ \mrm{a} = \lim_{h \to 0}\pars{1 + {h \over x}}^{x/h} = \lim_{n \to \pm\infty}\pars{1 + {1 \over n}}^{n} = \color{#f00}{\expo{}} \quad\mbox{which yields}\quad \totald{\log_{\,\mrm{a}}\pars{x}}{x} = {1 \over x} $$