Why is this for function specific range tending towards $e$

Question

I was messing around with the binomial approximation method as per which $(1+x)^{n} ≈ (1+nx)$ for $x<<1$, so while entering values in the calculator I observed something strange that the value, $(1+10^{-(n+1)})^{10^{n+1}}$ was tending towards e for some specific values, which sounded like a coincidence for 1 or 2 values. Still, roughly the function evaluated at a particular range giving roughly the exact value is absurdly not a coincidence. This got me here, as I'm confused about what's happening. I can't find any approximation like this on the internet.

Answer 1

Depending on what source you're using, sometimes $e$ is defined as $$e=\lim_{n\to\infty}\Bigl(1+\frac{1}{n}\Bigr)^n.$$ Of course, we have to show that this sequence has a limit, but then we can define $e$ to be whatever that limit is and deduce all kinds of other properties. Then this makes what you noticed unsurprising, because it is approximating the limit.

However, there are other ways to define $e$, which lead to more interesting explanations for what you observed. Define $f:\mathbb{R}\to (0,\infty)$ as $$f(x)=\sum_{n=0}^\infty \frac{x^n}{n!}.$$ One can show that this function is strictly positive, differentiable (and therefore continuous), $f'=f$, $f(0)=1$, $\lim_{x\to -\infty}f(x)=0$, and $\lim_{x\to\infty}f(x)=\infty$. Since $f$ is strictly increasing, it's one-to-one. Since it's continuous and approaches $0$ and $\infty$ as $-\infty$ and $\infty$, respectively, the intermediate value theorem says $f$ is onto $(0,\infty)$. Therefore there is an inverse function $g:(0,\infty)\to \mathbb{R}$ satisfying $$x=g(f(x))$$ for all $x\in \mathbb{R}$. Hitting both sides with $\frac{d}{dx}$ gives $$1=g'(f(x))f'(x)=g'(f(x))f(x),$$ and $$g'(f(x))=\frac{1}{f(x)}.$$ In other words, $g'(y)=1/y$ for all $y>0$. By the chain rule, $[g(h(x))]'=\frac{h'(x)}{h(x)}$ for any differentiable function $h:I\to (0,\infty)$.

We can also derive some properties. Fix $x'>0$ and define $h:(0,\infty)\to (0,\infty)$ by $h(x)=xx'$. Then $h'(x)=x'$, and $$[g(h(x))]'=\frac{h'(x)}{h(x)}=\frac{x'}{xx'}=\frac{1}{x}.$$ This means $g$, $g\circ h$ have the same derivative, and differ by a constant. Since $f(0)=1$, $g(1)=0$ and $g\circ h(1)=g(x')$. Therefore $g\circ h=g+g(x')$. Thus we have derived $$g(xx')=g(x)+g(x')$$ for all $x,x'>0$.

For $r\in \mathbb{R}$, let $h:(0,\infty)\to (0,\infty)$ be given by $h(x)=x^r$. Then $h'(x)=rx^{r-1}$ and $$[g(h(x))]'=\frac{h'(x)}{h(x)}=\frac{rx^{r-1}}{x^r}=\frac{r}{x}=rh'(x).$$ This means $g\circ h = rh+C$ for some constant $C$. Evaluating both sides at $x=1$ gives $C=0$. And we derived $g(x^r)=rg(x)$.

Now, rather than defining $$e=\lim_{x\to \infty}\Bigl(1+\frac{1}{x}\Bigr)^x,$$ we've defined $e=f(1)$, which is equivalent to saying that $e$ is the unique solution to $g(x)=1$.

Now let's evaluate \begin{align*} \underset{x\to \infty}{\lim} g\Bigl(\bigl(1+\frac{1}{x}\bigr)^x\Bigr) & = \underset{x\to\infty}{\lim} xg\Bigl(1+\frac{1}{x}\Bigr) = \underset{x\to\infty}{\lim}\frac{g\Bigl(1+\frac{1}{x}\Bigr)}{1/x} \\ & \overset{\frac{0}{0}}{=} \underset{x\to\infty}{\lim} \frac{g'\Bigl(1+\frac{1}{x}\Bigr)(-1/x^2)}{-1/x^2} =\underset{x\to\infty}{\lim}\frac{1}{1+\frac{1}{x}}=1.\end{align*}

Then by continuity of $f$, \begin{align*} \underset{x\to\infty}{\lim} \Bigl(1+\frac{1}{x}\Bigr)^x & = \underset{x\to\infty}{\lim} f\circ g\Bigl(\bigl(1+\frac{1}{x}\bigr)^x\Bigr) = f\Bigl(\underset{x\to\infty}{\lim} g\Bigl(\bigl(1+\frac{1}{x}\bigr)^x\Bigr)\Bigr) = f(1)=e.\end{align*}