justifying a definition of $e^x$

Question

I'm studying the different definitions of $e^x$ and showing their equivalences. If we define $e^x$ as the limit of the sequences of functions $f_n=(1+x/n)^n$, for all real $x$, how do we go about first justifying this limit exists for all reals? I can show $f_n$ is continuous on all of $\mathbb{R}$ and the sequence converges pointwise on $[0,1]$ but I haven't figured out how to carry on all of $\mathbb{R}$ and uniform convergence on compact intervals. Am I just supposed to show it converges to the series we all know and love and then carry on from there? Is there a way to do this without relying on the properties of other definitions?

Here is a sketch of some work,

Trivially, $\lim_{n\to \infty} f_n(0)=1$ and by the monotone convergence theorem, $\lim_{n\to \infty} (1+1/n)^n$ converges, and we denote its limit by $e$. Next we see $\lim_{n\to \infty} (1-1/n)^n=1/e$. Using the MCT one can show pointwise convergence on $(0,1)$. Now I've seen this argument before

For $x\neq 0$, we may write $$\lim_{n\to \infty} (1+x/n)^n=\lim_{n\to \infty} \left[\left(1+\frac{1}{n/x}\right)^{n/x}\right]^x=e^x$$

Why does $\left(1+\frac{1}{n/x} \right)^{n/x}\to e$? Is this assuming any other definition? What are some suggestions for other ways to show this limit exists?

Answer 1

The thing to prove would by $$\lim_{x\to\infty}\left(1+\frac{1}{x}\right)^x=e$$ where $x$ is a real number.

To do this, just notice that if $n\leq x<n+1$ for some natural $n$, then $$\left(1+\frac{1}{n}\right)^n\leq\left(1+\frac{1}{x}\right)^x\leq\left(1+\frac{1}{n+1}\right)^{n+1}$$

Answer 2

You need to study the related sequence $g_{n}(x) = \left(1 - \dfrac{x}{n}\right)^{-n}$ also. For $x \geq 0$ it is easy to show (via binomial theorem for positive integer index) that $f_{n}(x)$ is increasing function of $n$ $$1 \leq f_{n}(x) \leq 1 + x + \frac{x^{2}}{2!} + \dots + \frac{x^{n}}{n!}$$ Since the series on the right converges (say to sum $E(x)$) as $n \to \infty$ it follows that $$f_{n}(x) \leq E(x)$$ for all $x \geq 0$. It follows that for $x \geq 0$, the sequence $f_{n}(x)$ is increasing and bounded and hence convergent.

For $x < 0$ we need some new ideas.

First of all we can prove very easily that for all $x$ we have $$f_{n}(x)/g_{n}(x) \to 1$$ as $n \to \infty$. So $g_{n}(x)$ also tends to same limit as $f_{n}(x)$ (if it exists). Since $f_{n}(x)$ tends to a limit (say $f(x)$ and $f(x) \geq 1$) if $x \geq 0$, it follows that $g_{n}(x)$ also tends to the same limit when $x \geq 0$. Now we have $f_{n}(-x) = 1/g_{n}(x)$ and hence for $x \geq 0$ we have $f_{n}(-x) \to 1/f(x)$. It follows that $f_{n}(x)$ converges for $x < 0$ also. And Since $f_{n}(-x)g_{n}(x) = 1$ it follows that $f(-x)f(x) = 1$ for all $x$.

You can continue the theory by proving further that $f(x + y) = f(x)f(y)$ for all $x, y$.