Finding the limit: $\lim_{x\to \infty}\frac{1}{2}x\sin {\frac{180(x-2)}{x}}$

Question

While investigating a problem, I came across a function: $$f(x) = \frac{1}{2}x\sin {\frac{180(x-2)}{x}}$$ When looking at the function in Desmos (I was checking my proof), I discovered that $$\lim_{x\to \infty}\frac{1}{2}x\sin {\frac{180(x-2)}{x}} = \pi$$ I double-checked Wolframalpha, and this limit is true. The only issue is that I can't seem to prove it by hand, and I'm really interested as to how $\pi$ pops out of nowhere. Please note I am working in degrees, so it is not 180 radians in the sin function. I would really appreciate it if someone could explain a solution.

Answer 1

You really shouldn't work in degrees; more specifically, the sin function itself is defined (despite what you might have learned in high school) with 'radian' arguments. Formulae like $e^{ix}=\cos x+i\sin x$, or $\frac{d}{dx}\sin x=\cos x$, rely on it. There's another formula that relies on it that's the critical one here: $\lim_{x\to 0}\frac{\sin x}x=1$. (Incidentally, another way of thinking about this might be that since angles are 'dimensionless', unlike length or spans of time or mass, $180^\circ$ is literally just a fancy way of writing '$\pi$')

Now, writing your function 'properly', you have $f(x)=\frac12x\sin\left(\pi(1-\frac2x)\right)$ $= \frac12 x\sin(\pi-\frac{2\pi}{x})$. Using the symmetry of the $\sin$ function, this is equal to $\frac12x\sin(\frac{2\pi}x)$. Now, we can substitute $y=\frac1x$; taking the limit as $x\to\infty$ is the same as taking the limit as $y\to 0$ (technically only from positive $y$, but that's moot here), and so your limit is equal to $\frac12\lim_{y\to 0}\dfrac{\sin(2\pi y)}{y}$. But $\lim_{y\to 0}\frac{\sin(ay)}y$ $= a\lim_{y\to 0}\frac{\sin(ay)}{ay}$ $=a$; this gives your limit as $\frac12\cdot2\pi=\pi$.

Answer 2

By assumption that your $180$ it's $180^{\circ}$ we obtain: $$f(x)=\frac{1}{2}x\sin\frac{2\pi}{x}=\frac{\sin\frac{2\pi}{x}}{\frac{2\pi}{x}}\cdot\pi\rightarrow\pi.$$

Answer 3

As an alternative, by l'Hopital we have that

$$\lim_{x\to \infty }\frac{1}{2}x\sin {\frac{\pi(x-2)}{x}}=\lim_{x\to \infty }\frac12 \frac{\sin {\left(\pi-\frac{2\pi}{x}\right)}}{\frac1x}=$$

$$=\lim_{x\to \infty }\frac12 \frac{\frac{2\pi}{x^2}\cos {\left(\pi-\frac{2\pi}{x}\right)}}{-\frac1{x^2}}=\frac12\frac{2\pi(-1)}{(-1)}=\pi$$