I will show that
the difference between
the areas of the
circumscribed and inscribed
$n$-gons goes to zero.
Consider a regular $n$-gon
inscribed in the unit circle.
There are $2n$ triangles
with central angle
$t = \pi/n$
and hypotenuse
$1$,
so the distance to the side
$s_n$ and length $h_n$
satisfy
$s_n = \cos(t)$
and $h_n = \sin(t)$.
The area of each triangle
is thus
$\frac12 s_nh_n
=\frac12\cos(t)\sin(t)
=\frac14\sin(2t)
$
so the area of the
inscribed $n$-gon is
$2n$ times this or
$\frac12n\sin(2t)
$.
Extend the radii
to get the
circumscribed $n$-gon.
There are $2n$ triangles
with base $1$
and height
$g_n$ such that
$g_n = \tan(t)$,
so the area is
$\frac12 g_n
=\frac12 \tan(t)
$.
The total area is thus
$n\tan(t)
$.
Note that
both of these areas go to
$\pi$ as $n \to \infty$
since
$\sin(x) \approx \tan(x)
\approx x$
as $x \to 0$.
However,
the only inequality needed is
$\sin(x) < x$
for $0 < x < \pi/2$.
The difference in
the two areas is thus
$\begin{array}\\
D_n
&=n\tan(t)-\frac12 n\sin(2t)\\
&=n\left(\tan(t)-\frac12 \sin(2t)\right)\\
&=n\left(\dfrac{\sin(t)}{\cos(t)}-\sin(t)\cos(t)\right)\\
&=n\sin(t)\left(\dfrac{1}{\cos(t)}-\cos(t)\right)\\
&=n\sin(t)\dfrac{1-\cos^2(t)}{\cos(t)}\\
&=n\sin(t)\dfrac{\sin^2(t)}{\cos(t)}\\
&=n\dfrac{\sin^3(t)}{\cos(t)}\\
&=n\dfrac{\sin^3(t)}{1-2\sin^2(t/2)}\\
\end{array}
$
We now use
$\sin(x) < x$
for
$0 < x < \pi/2$
so
$\begin{array}\\
D_n
&\lt n\dfrac{t^3}{1-2(t/2)^2}\\
&= n\dfrac{(\pi/n)^3}{1-2(\pi/(2n))^2}\\
&= \dfrac{1}{n^2}\dfrac{\pi^3}{1-\pi^2/(2n^2)}\\
&< \dfrac{2\pi^3}{n^2}
\qquad\text{for } n \ge 4\\
&\to 0
\qquad\text{as } n \to \infty\\
\end{array}
$