It's All Tangents!

$\lim_{n \rightarrow \infty} \sum_{j = 1}^{n} (\dfrac{j}{n})^n x^{n - j} = \sum_{j = 0}^{n - 1} (1 - \dfrac{j}{n})^{n} x^{j} = \sum_{n = 0}^{\infty} (\dfrac{x}{e})^n = \dfrac{e}{e - x}$ on $|x| < e$

$\implies f(x) = \lim_{n \rightarrow \infty} \dfrac{\sum_{j = 1}^{n} x^{j}}{\sum_{j = 1}^{n} (\dfrac{j}{n})^n x^{n - j}} = (\dfrac{x}{1 - x})(\dfrac{e - x}{e}) = \dfrac{x(e - x)}{e(1 - x)}$ on $|x| < 1$

$\implies g(x) = \dfrac{\sum_{n = 1}^{\infty} (-1)^{n + 1} x^n}{\sum_{n = 0}^{\infty} (-1)^n (\dfrac{x}{e})^n} =$ $\dfrac{-\sum_{n = 1}^{\infty} (-x)^n}{\sum_{n = 0}^{\infty} (\dfrac{-x}{e})^n} =$ $\dfrac{x(e + x)}{e(1 + x)}$ on $|x| < 1$ .

$\boxed{f(x) = \dfrac{x(e - x)}{e(1 - x)}}$ and $\boxed{g(x) = \dfrac{x(e + x)}{e(1 + x)}}$ on $|x| < 1$ .

and

$\dfrac{d}{dx}(f(x)) = \dfrac{x^2 - 2x + e}{e(1 - x)^2}$ and $\dfrac{d}{dx}(g(x)) = \dfrac{x^2 + 2x + e}{e(1 + x)^2}$

and,

$\dfrac{d}{dx}(g(x))|_{x = a} = \dfrac{a^2 + 2a + e}{e(1 + a)^2} = \dfrac{d}{dx}(f(x))|_{x = -a}$ .

Let $a = e - 2 \implies \dfrac{d}{dx}(g(x))|_{x = e - 2} = \dfrac{1}{e - 1} = \dfrac{d}{dx}(f(x))|_{x = 2 - e}$

$A: (e - 2,\dfrac{2(e - 2)}{e})$ and $B: (2 - e,\dfrac{2(2 - e)}{e})$

Using point $B: (2 - e,\dfrac{2(2 - e)}{e}) \implies e(e - 1)y - ex = -(e - 2)^2$ and the normal line to $f(x)$ at $x = e - 2$ with slope $m_{\perp} = -(e - 1)$ is: $\boxed{ey + e(e - 1)x = -(e - 2)(e^2 - e + 1)}$

Using point $A: (e - 2,\dfrac{2(e - 2)}{e}) \implies \boxed{e(e - 1)y - ex = (e - 2)^2}$

$\boxed{ey + e(e - 1)x = -(e - 2)(e^2 - e + 1)}$

$\boxed{e(e - 1)y - ex = (e - 2)^2}$

Solving the system above we obtain:

$x_{0} = -\dfrac{e^4 - 4e^3 + 8e^2 - 12e + 8}{e(e^2 - 2e + 2)}$ and $y_{0} = \dfrac{2(2 - e)}{e^2 - 2e + 2}$ .

Using $C: (x_{0},y_{0})$ and $B: (2 - e,\dfrac{2(2 - e)}{e}) \implies BC = \dfrac{2(e - 2)^2}{e\sqrt{e^2 - 2e + 2}} \approx \boxed{0.190936}$ .