Why does $y = x\sin(\frac{180}{x})$ approach $\pi$? A few days ago I was playing on my scientific calculator and I ran over an interesting little equation: $180\sin(1)$ is extremely close to $\pi$. At first I thought it was a coincidence, but then I tried $360\sin\left(\frac{1}{2}\right)$ and it was closer to pi. So then I tried out $90\sin(2)$ and it was farther from $\pi$. So I came up with the equation $y = 180x\sin\left(\frac{1}{x}\right)$, which I then simplified to $y = x\sin\left(\frac{180}{x}\right)$. Although it approaches $\pi$ slower (180 times slower), it is easier to understand what was going on on the desmos graphing calculator. It seems the larger the $x$ value, the closer it gets to $\pi$. I would like to know why exactly this happens.

EDIT: (Sine is in degrees not radians, just for clarification)
Given that you are still using degrees for trig functions, I am not sure how familiar you are with calculus. But I will try to explain what's happening without  calculus, and then show what happens with calculus. 

As $x$ gets larger, it will cause your equation $y$ to increase. At the same time, the argument inside of $\sin$ is getting very small because you are dividing by $x$. When $\sin$ is evaluated near zero, the output will also be very near zero. You may have seen before that $$\sin(\theta) \approx \theta$$ when $\theta$ is small. Anyway, long story short, you are observing a mathematical battle where one part of the functions wants to diverge to infinity, and it is being multiplied by another part of the function that is tending toward zero. As you have noticed, the battle will balance out at $\pi$ in this particular case. Once you know (more) calculus, you will have tools to measure just how fast functions are growing or shrinking, and can evaluate limits like these. One useful tool is known as "L'Hospital's Rule" , which when applied to your function would say that $$lim_{x \rightarrow \infty} \left[x \cdot \sin\left(\frac{180}{x}\right)\right]$$ $$=lim_{x \rightarrow \infty} \left[\frac{ \sin\left(\frac{180}{x}\right)}{x^{-1}}\right]$$ $$=lim_{x \rightarrow \infty} \left[\frac{ \cos\left(\frac{180}{x}\right)\cdot \frac{-180}{x^2}}{-x^{-2}}\right]$$ $$=lim_{x \rightarrow \infty} \left[180\cos\left(\frac{180}{x}\right)\right]$$ Now you can argue that since $lim_{x \rightarrow \infty}\frac{180}{x}=0$ and $\cos(0)=1$ that $lim_{x \rightarrow \infty}\cos\left(\frac{180}{x}\right)=1$. So $$lim_{x \rightarrow \infty} \left[180\cos\left(\frac{180}{x}\right)\right]=180 \cdot 1=180$$ And since $180$ degrees is the same as $\pi$ radians, we have shown mathematically just what you have witnessed on your calculator.

EDIT: @Rodolvertice If you are interested in more limits like these, I will say that I think a very cool result comes from $$\lim_{x \rightarrow \infty} \left[\left(1+\frac{1}{x} \right)^x\right]$$
So I came to the same conclusion once, that as "x" approaches infinity, then the value "xSin(180/x)", in degrees, approaches pi. But I had a more deliberate approach at this the first time I figured it out.

It's quite simple... I treated every other regular polygon as we treat a circle... so that they too would have constants, as "pi" is to a circle... deriving a formula that would relate the number of sides of the polygon "n" and the longest distance from the center of the polygon "D" to the polygon's perimeter "C".Just as is "C = (pi)D". 

I then ended up with "C or P = [nSin(180/n)]D". So that the constant of the shape "k" = "nSin(180/n)". I thought that, by logic, this SHOULD approach pi as "n" approaches infinity.
It gave me this graph enter image description here

So you can see that when you calculate "xSin(180/x)", your value of "x" is actually the number of sides of that regular shape... and as "x" approaches infinity, i.e. infinite sides of circle, you get (pi).

You could actually try and play around to get to the same conclusion I did. It's pretty correct. I hope this helps :)