The following function is half the perimeter of a regular $n$-gon with circumradius $1$ where $x$ is the number of subdivisions in half following a hexagon (when $x=1$, $n$ is $12$):
$p(x)=6 (2^{x-1})\sqrt{2-2\cos(\frac{\pi}{3(2^x)})}$
Geometrically and numerically, it is obvious that $\lim \limits_{x \to \infty} p(x) = \pi$.
How can you prove this algebraically?
As a bonus question, is there any closed form expression with no trig functions that approaches $\pi$? This came up in an attempt to find one. It failed, but I did find out an interesting recurrence relation:
$q(x)=\sqrt{2-2\cos(\frac{\pi}{3(2^x)})}$
satisfies:
$q(x+1)=\sqrt{2-\sqrt{4-q(x)^2}}$
at least when $x$ is a non-negative integer.
Let's do a couple algebraic manipulations to get: $p(x)=6\sqrt{2}\cdot 2^{x-1}\sqrt{1-\cos\left(2\cdot \frac\pi{12\cdot2^{x-1}}\right)}$
Now, if we let $y=2^{x-1}$, then $y\to\infty$ as $x\to\infty$ and so our limit is equal to $L=\lim_{y\to\infty}6\sqrt 2\cdot y\sqrt{1-\cos\left(2\cdot\frac\pi{12y}\right)}$
But $\cos(2\theta)=1-2\sin^2(\theta)$ and so $L=\lim_{y\to\infty}6\sqrt{2}\cdot y\cdot\sqrt{2\sin^2\left(\frac\pi{12y}\right)}=\lim_{y\to\infty}12y\sin\left(\frac\pi{12y}\right)$
One last substitution in case it isn't clear yet, $t=\frac1y$ so that $t\to0$ as $y\to\infty$:
$$L=12\lim_{t\to0}\frac{\sin\left(\frac\pi{12}t\right)}t=12\cdot\frac\pi{12}=\boxed{\pi}$$
As for your bonus question, it really does depend on what you mean by "closed form." I would argue that $\pi=\frac{\ln(-1)}i$ is a closed form expression. $\pi=2\displaystyle\int_{-1}^1\sqrt{1-t^2}dt$ is also perfectly valid and can be converted into a limit of a Riemann sum. Generally, however, the fundamental property of $\pi$ as $\frac1{2i}$ of the period of $\exp$ means that you'll probably want to invoke the complex exponential (or the trig functions that are just the complex exponential in disguise) in your "closed-form" limits for $\pi$.
Hope this helps!