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I have read proofs that the sequence $a_n=\left( 1+\frac{1}{n}\right)^n,n\in\mathbb{N^*}$ converges and its limit is defined to be $e$.

How is this definition of $e$ extended to the real numbers?

Phrased in another way: How can we prove the following:

Letting $f(x) = \left( 1+\frac{1}{x} \right) ^x$, prove that $\lim_\limits{x\to +\infty}{f(x)}=e$.

This has to be proven only with the definition that $e=\lim_\limits{n\to +\infty}{\left[ \left( 1+\frac{1}{n}\right)^n\right] }, n\in\mathbb{N^*}$, for the extension to be correct.

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    $\begingroup$ You can prove for instance that $\left(1+{1\over n}\right)^n <\left(1+{1\over x}\right)^x<\left(1+{1\over n+1}\right)^{n+1}$ if $n<x<n+1$. $\endgroup$ – Aretino Sep 18 '15 at 17:36
  • $\begingroup$ Adding to the comment by @Arentino, the inequality is evident since $f$ is monotonically increasing, which can be shown using standard (non-calculus) tools. $\endgroup$ – Mark Viola Sep 18 '15 at 17:41
  • $\begingroup$ Use the power series for log(1+y) for small |y| to show that the derivative of log(f(x)) is positive for positive x. $\endgroup$ – DanielWainfleet Sep 18 '15 at 23:07
  • $\begingroup$ @Aretino Can you post a mathematically complete proof as an answer? $\endgroup$ – Jason Sep 19 '15 at 6:05
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    $\begingroup$ See paramanands.blogspot.com/2014/05/… which uses the rigorous equivalent of your definition. $\endgroup$ – Paramanand Singh Sep 20 '15 at 4:17
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Let $n=\lfloor x\rfloor$, then: $$ \left(1+{1\over n+1}\right)^n \le \left(1+{1\over x}\right)^x \le \left(1+{1\over n}\right)^{n+1}, $$ that is $$ \tag{1} {\left(1+{1\over n+1}\right)^{n+1}\over \left(1+{1\over n+1}\right)} \le \left(1+{1\over x}\right)^x \le \left(1+{1\over n}\right)^{n}\left(1+{1\over n}\right). $$ The left and right hand side of this inequality can be viewed as two sequences, $a_n$ and $b_n$, both converging to $e$ (by definition of $e$ and standard theorems on the limit of a product of two sequences). So for any $\epsilon>0$ there exists $M$ such that $e-\epsilon<a_n$ and $b_n<e+\epsilon$ when $n>M$.

In virtue of $(1)$ we have then $e-\epsilon< \left(1+{1\over x}\right)^x <e+\epsilon$ for all $x>M+1$, that is: $\lim\limits_{x\to\infty}\left(1+{1\over x}\right)^x=e$.

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  • $\begingroup$ +1 for getting those bounds for $(1 + (1/x))^{x}$ I have mentioned a few lines about your approach in my answer. $\endgroup$ – Paramanand Singh Sep 20 '15 at 6:18
  • $\begingroup$ @ParamanandSingh I've read your answer but I think that monotonicity of $a^x$ with respect to base and exponent is automatically entailed by any reasonable definition of real exponent power. One usually starts with the definition for integer exponent, where monotonicity holds, and extends it to rational exponents, where it can be proved to still hold. In view of that a natural definition of power with real exponent $x$ can be given as $a^x=\sup_{y\in Q,y<x}a^y=\inf_{y\in Q,y>x}a^y$, which is the definition given by the OP. From that it is straightforward to prove that monotonicity holds. $\endgroup$ – Aretino Sep 20 '15 at 8:07
  • $\begingroup$ Oh yes for proving monotone nature of $a^{x}$ you don't need the derivative. But for $x^{a}$ you do need it. $\endgroup$ – Paramanand Singh Sep 20 '15 at 10:41
  • $\begingroup$ @ParamanandSingh I think one can do without derivatives. If $1<x<y$ then you can prove $x^q<y^q$ for any rational $q$. Given $a\in\mathbb{R}$ just take a sequence $q_n$ of rational numbers converging to $a$: one has $x^{q_n}<y^{q_n}$ for any $n$ and passing to the limit that gives $x^a\le y^a$. Strict inequality can be obtained by proving $x^a$ to be injective, but that is not necessary here. $\endgroup$ – Aretino Sep 20 '15 at 11:26
  • $\begingroup$ Agree. I will update my answer. $\endgroup$ – Paramanand Singh Sep 20 '15 at 12:36
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I am trying to summarize the essential ideas needed to establish $$\lim_{x \to \infty}\left(1 + \frac{1}{x}\right)^{x} = e$$ based on the definition of $a^{x}$ which OP has provided in his comments.

We need to establish the following results in order:

1) $a^{x}a^{y} = a^{x + y}$

2) $a^{x}$ is strictly monotone for $a > 0, a \neq 1$. For $0 < a < 1$ it is decreasing and for $a > 1$ it is increasing as a function of $x$.

3) $a^{x} \to 1$ as $x \to 0$. This (combined with point 1) above) establishes that $a^{x}$ is continuous for all $x$.

4) $\lim_{x \to 0}\dfrac{a^{x} - 1}{x}$ exists for all $a > 0$ and hence defines a function $L(a)$ for all $a > 0$. This also establishes that $a^{x}$ is differentiable with $(a^{x})' = a^{x}L(a)$.

4a) $L(1) = 0, L(ab) = L(a) + L(b)$ and $L(a)$ is a strictly increasing function of $a$.

4b) There is a unique number $\xi > 0$ such that $L(\xi) = 1$.

4c) $L(a^{x}) = xL(a)$

4d) $\lim_{x \to 0}\dfrac{L(1 + x)}{x} = 1$ Combined with 4a) this shows that $L(x)$ is continuous/differentiable for all $x > 0$ with $L'(x) = 1/x$.

And finally using 4c) and 4d) we show that $$L\left\{\left(1 + \frac{1}{x}\right)^{x}\right\} \to 1$$ as $x \to \infty$. By continuity and monotone nature of $L$ it follows that $(1 + (1/x))^{x} \to \xi$ where $L(\xi) = 1$. The limit remains same if $x$ tends to $\infty$ through integer values and hence $$\lim_{x \to \infty}\left(1 + \frac{1}{x}\right)^{x} = \lim_{n \to \infty}\left(1 + \frac{1}{n}\right)^{n} = \xi$$ This $\xi$ is traditionally denoted by $e$ and $L(x)$ is denoted by $\log x$.

All this has been developed in detail in my blog post. The difficult part is to establish 4) and its subpart 4d).

The approach by Aretino is very smart in the sense that it uses the monotone nature of both functions $x^{a}$ and $a^{x}$ to bound the function $(1 + (1/x))^{x}$.

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