Common Tangents, Areas, Logs and Limits.

Let $f(x) = x^{2n + 1}$ and $g(x) = \log_{b}(x) = \dfrac{\ln(x)}{\ln(b)}$ for $x > 0$

$\implies f'(a) = (2n + 1)a^{2n} = g'(a) = \dfrac{1}{a\ln(b)} \implies$ $a^{2n + 1} = \dfrac{1}{(2n + 1)\ln(b)} \implies a = (\dfrac{1}{(2n + 1)\ln(b)})^{\frac{1}{2n + 1}}$

and

$a^{2n + 1} = \dfrac{\ln(a)}{\ln(b)} \implies \dfrac{1}{(2n + 1)\ln(b)} = \dfrac{\ln(\dfrac{1}{(2n + 1)\ln(b)})}{(2n + 1)\ln(b)} \implies \ln(\dfrac{1}{(2n + 1)\ln(b)}) = 1 \implies \ln(b) = \dfrac{1}{2n + 1)e} \implies$

$\large b = e^{\frac{1}{(2n + 1)e}} \implies a = e^{\frac{1}{2n + 1}} \implies a^{2n + 1} = e$

$\implies \large A:(e^{\frac{1}{2n + 1}}, e)$ and using the symmetry about the origin we have $\large A':(-e^{\frac{1}{2n + 1}}, -e)$

$\implies m_{AA'} = \large e^{\frac{2n}{2n + 1}} \implies y = e^{\frac{2n}{2n + 1}}x$

$\implies A_{n} = \large 2\displaystyle\int_{0}^{e^{\frac{1}{2n + 1}}} (e^{\frac{2n}{2n + 1}}x - x^{2n + 1}) \:\ dx = (\dfrac{n}{n + 1})e^{\frac{2n + 2}{2n + 1}}$

$\implies \large \lim_{n \rightarrow \infty} A_{n} = e\lim_{n \rightarrow \infty} (1 - \dfrac{1}{n + 1})e^{\frac{1}{2n + 1}} = e$ $\approx \boxed{2.7182818}$ .