Definition extension of $e$ to the reals

Question

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.

Answer 1

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: $\lim\limits_{x\to\infty}\left(1+{1\over x}\right)^x=e$.

Answer 2

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 $(1 + (1/x))^{x}$.