Tangent Craze

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

To show $y = x$ is parallel to $\overleftrightarrow{AB}$ .

$\dfrac{d}{dx}(f(x))|_{x = a} = 1 \implies a = \dfrac{-1}{(2n)^{\frac{1}{2n - 1}}}$ and $\dfrac{d}{dx}(g(x))|_{x = b} = 1 \implies b = \dfrac{1}{(2n)^{\frac{2n}{2n - 1}}}$

$\implies A: (\dfrac{-1}{(2n)^{\frac{1}{2n - 1}}},\dfrac{-1}{(2n)^{\frac{2n}{2n - 1}}})$ and $B: (\dfrac{1}{(2n)^{\frac{2n}{2n - 1}}},\dfrac{1}{(2n)^{\frac{1}{2n - 1}}}) \implies m_{AB} = 1 \implies \overleftrightarrow{AB}$ is parallel to line $y = x$

and $y = x + \dfrac{2n - 1}{(2n)^{\frac{2n}{2n - 1}}}$ .

Using the symmetry about the $y$ axis $\implies A^{'}: (\dfrac{1}{(2n)^{\frac{1}{2n - 1}}},\dfrac{-1}{(2n)^{\frac{2n}{2n - 1}}})$ and $B^{'}: (\dfrac{-1}{(2n)^{\frac{2n}{2n - 1}}},\dfrac{1}{(2n)^{\frac{1}{2n - 1}}})$ $\implies m_{A^{'}B^{'}} = -1 \implies y = -x + \dfrac{2n - 1}{(2n)^{\frac{2n}{2n - 1}}}$ .

$BB^{'} = \dfrac{2}{(2n)^{\frac{2n}{2n - 1}}}, AA^{'} = \dfrac{2}{(2n)^{\frac{1}{2n - 1}}}$ , and $P: (\dfrac{-1}{(2n)^{\frac{2n}{2n - 1}}}, \dfrac{-1}{(2n)^{\frac{2n}{2n - 1}}}) \implies$

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

$\implies A_{n} = \dfrac{1}{2}(AA^{'} + BB^{'})(B^{'}P) = (\dfrac{2n + 1}{(2n)^{\frac{2n}{2n - 1}}})^2$

and $\lim_{n \rightarrow \infty} A_{n} = \lim_{n \rightarrow \infty} ((\dfrac{1}{2n})^{\frac{1}{2n - 1}} (1 + \dfrac{1}{2n}))^2 = \boxed{1}$ = $A_{\triangle{TQR}}$