Shouldn't $2^x \sin\frac{180}{2^x}$ approach $\pi$ as $x$ gets large? Take a  circle with diameter $1$. Obviously its circumference is $\pi$.
Draw a square inside this circle (biggest such). Since the circle's diameter is 1, we can work out that each side of the square would be $\frac{1}{\sqrt{2}}$. Its perimeter, therefore, is $2\sqrt{2}$.
Imagine a triangle being drawn over each side of the square, so as to create a regular octagon. Using the law of sines, we can work out the measure of each side of the octagon. The base of each triangle would be $\frac{1}{\sqrt{2}}$, and the angle opposite to it $135^o$. Each of the other angles would be $\frac{180-135}{2} = 22.5^o$.
So,
$$
\frac{1}{\sin135(\sqrt2)} = \frac{x}{\sin22.5}
$$
$$
1 = \frac{x}{\sin22.5}
$$
$$
x=\sin22.5
$$
Therefore the perimeter of this octagon is $8\sin22.5$.
We can continue drawing triangles on this octagon, and more triangles on that, and so on. Once the number of sides reaches infinity (which, of course, is not possible in the physical world), the perimeter of the said figure would be $\pi$.
I am trying to write a similar function $f(x)$ such that
$$
\lim_{x \to 0}f(x) = \pi
$$
Note that $f(x)$ is a function of levels. Level 1 is for 4 sides, level 2 is for 8 sides, then 16 sides, 32 sides and so on.
Insofar:
$$f(1) = 2\sqrt2 = 4 \sin 45$$
$$f(2) = 8 \sin22.5$$
$$...$$
I work out that 

$$f(x) = 2^x \sin\frac{180}{2^x}$$

When graphing that function:

Given this, it doesn't seem to limit on the graph. Say, if it does limit to $\pi$, would it approximate $\pi$ for a relatively large $x$?
Please note. There might be a lot of mistakes here. I'm not studying math at the college-level and is a careless person.

The limit should be $180$ if you are taking the sine in radians.  Yes, it should be $\pi$ if you are taking the sine in degrees.  You should be graphing for larger values of $x$ as even $\frac {180}{2^8} \gt \frac 12$ is not so small.  Below is a plot from Alpha that shows nice convergence to $180$ using radians

