Common Tangent Lines and Areas.

Let $f(x) = x^{\frac{1}{2n + 1}}$ and $g(x) = x^{2n + 1}$

First we find the area $A(n)$ of the regions $R_{1}$ and $R_{2}$ above.

From the graph above $R_{(0,0)} \circ R_{y = x} (x_{0},y_{0}) = (-y_{0},-x_{0})$

Using $A: (x_{0},y_{0})$ and $B: (-y_{0},-x_{0}) \implies$ slope $m_{AB} = 1 \implies$

$f'(x_{0}) = \dfrac{1}{2n + 1}(\dfrac{1}{x_{0}})^{\frac{2n}{2n + 1}} = 1 \implies$ $x_{0} = (\dfrac{1}{2n + 1})^{\frac{2n + 1}{2n}} \implies$

$A: ((\dfrac{1}{2n + 1})^{\frac{2n + 1}{2n}} , (\dfrac{1}{2n + 1})^{\frac{1}{2n}})$ and $B: (-(\dfrac{1}{2n + 1})^{\frac{1}{2n}}),-(\dfrac{1}{2n + 1})^{\frac{2n + 1}{2n}})$

$\implies y = x + (2n)j(n)$ , where $j(n) = (\dfrac{1}{2n + 1})^{\frac{2n + 1}{2n}}$

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

$\dfrac{1}{2}x^2 + (2n)j(n)x + (\dfrac{2n + 1}{2n + 2})x^{\frac{2n + 2}{2n + 1}}|_{0}^{j(n)} =$

$(\dfrac{4n + 1}{2})(j(n))^2 - (\dfrac{2n + 1}{2n + 2})(j(n))^{\frac{2n + 2}{2n + 1}} =$

$(\dfrac{4n + 1}{2})(\dfrac{1}{2n + 1})^{\frac{2n + 1}{n}} - (\dfrac{2n + 1}{2n + 2})(\dfrac{1}{2n + 1})^{\frac{n + 1}{n}} =$

$\boxed{\dfrac{1}{2}(\dfrac{n}{n + 1})(\dfrac{1}{2n + 1})^{\frac{2n + 1}{n}}}$

and

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

$\dfrac{1}{2}x^2 + (2n)j(n)x - (\dfrac{1}{2n + 2})x^{2n + 2}|_{-(\dfrac{1}{2n + 1})^{\frac{1}{2n}}}^{0} =$

$-(\dfrac{1}{2}(\dfrac{1}{2n + 1})^{\frac{1}{n}} - (2n)(\dfrac{1}{2n + 1})^{\frac{n + 1}{n}} - (\dfrac{1}{2n + 2})(\dfrac{1}{2n + 1})^{\frac{n + 1}{n}}) =$

$-(\dfrac{1}{2}(\dfrac{1}{2n + 1})^{\frac{1}{n}} - (\dfrac{1}{2n + 1})^{\frac{n + 1}{n}} (\dfrac{(2n + 1)^2}{2(n + 1)})) =$

$\boxed{\dfrac{1}{2}(\dfrac{n}{n + 1})(\dfrac{1}{2n + 1})^{\frac{1}{n}}}$

$\implies A_{R_{1}}(n) = A_{1}(n) + A_{2}(n) = \boxed{(\dfrac{n}{n + 1})(\dfrac{2n^2 + 2n + 1}{(2n + 1)^2})(\dfrac{1}{2n + 1})^{\frac{1}{n}}} = A_{R_{2}}(n)$

$\implies \lim_{n \rightarrow \infty} A_{R_{1}}(n) = \dfrac{1}{2}\lim_{n \rightarrow \infty} (\dfrac{1}{2n + 1})^{\frac{1}{n}}$

and

$\lim_{n \rightarrow \infty} (\dfrac{1}{2n + 1})^{\frac{1}{n}} =$ $\lim_{n \rightarrow \infty} e^{(\dfrac{\ln(\frac{1}{2n + 1})}{n})} =$

$e^{\lim_{n \rightarrow \infty} (\dfrac{\ln(\frac{1}{2n + 1})}{n})} =$ $e^{\lim_{n \rightarrow \infty} (\dfrac{-2}{2n + 1})} = e^{0} = 1$

$\implies \lim_{n \rightarrow \infty} A_{R_{1}}(n) = \dfrac{1}{2} \implies$

Total Area $\boxed{A = \lim_{n \rightarrow \infty} A(n) = 2\lim_{n \rightarrow \infty} A_{R_{1}}(n) = 1}$ .

Next we find the rectangular area $A_{ABDC}(n)$ .

Using the coordinates of points $A,B,C$ above and point $D: (-x_{0},-y_{0})$ we have:

$AC = \sqrt{2}|y_{0} - x_{0}|$ and $AB = \sqrt{2}|y_{0} + x_{0}| \implies$

$A_{ABDC}(n) = 2|y_{0}^2 - x_{0}^2| = 2|(\dfrac{1}{2n + 1})^{\frac{1}{n}} - (\dfrac{1}{2n + 1})^{\frac{2n + 1}{n}}| =$

$2(\dfrac{1}{2n + 1})^{\frac{1}{n}}(1 - (\dfrac{1}{2n + 1})^2) =$ $\boxed{\dfrac{2(4n)(n + 1)}{(2n + 1)^2}(\dfrac{1}{2n + 1})^{\frac{1}{n}}}$ .

From above $\lim_{n \rightarrow \infty} (\dfrac{1}{2n + 1})^{\frac{1}{n}} = 1 \implies \boxed{A^{*} = \lim_{n \rightarrow \infty} A_{ABDC}(n) = 2}$

$\implies \lim_{n \rightarrow \infty} A_{T}(n) = A^{*} - A = \boxed{1}$

From the picture, we see that the points A,B, C and D are where the tangent is 1. Also noted that f and g are inverses of the others.

By thinking about the shape of the graph when n is large, one can see that the limit is an area of 2 triangles with vertices (0,0), (1,0), (0,1), (-1,0) and (0,-1), so, the area is then $\boxed{1}$