Why does $\pi$ turn out to be negative

Question

I did the following to derive the value of $\pi$, you might want to grab a pencil and a piece of paper:

Imagine a unit circle with center point $b$ and two points $a$ and $c$ on the circumference of the circle such that triangle $abc$ is an obtuse triangle. you can see that if $\theta$ denotes the angle $\angle acb$ then $0<\theta<90$ and that the angle of the sector $abc$ is $180 -2\theta$, so the area of sector $abc$ is $\frac{180-2\theta}{360}\pi = \frac{90-\theta}{180}\pi$. If we extend the radius $bc$ to form a diameter $D$, then the angle between line $ab$ and $D$ is $180-(180-2\theta) = 2\theta$; so if we define the distance between the point $a$ and the line $D$ as $h$ we get $h = \sin(2\theta)$. These allows us to derive the area of triangle $abc$ as $\frac{1}{2}sin(2\theta)$. The area of segment $ac$ is the area of sector abc - the area of triangle abc : $$ \frac{90-\theta}{180}\pi - \frac{1}{2}\sin(2\theta) $$ We can see that as $\theta$ approaches 0, the area of segment ac approaches the half the area of the circle which is $\frac{\pi}{2}$ $$ \lim_{\theta \to 0} \frac{90-\theta}{180}\pi - \frac{1}{2}\sin(2\theta) = \frac{\pi}{2} $$ $$ \lim_{\theta \to 0} \frac{90-\theta}{90}\pi - \sin(2\theta) = \pi $$ $$ \lim_{\theta \to 0} \pi\Big[\frac{90-\theta}{90} - 1\Big] = \lim_{\theta \to 0} \sin(2\theta) $$ $$ \lim_{\theta \to 0} -\frac{\theta\pi}{90} = \lim_{\theta \to 0} \sin(2\theta) $$ $$ \pi = -\lim_{\theta \to 0} \frac{90\sin(2\theta)}{\theta} $$ However this limit approaches -3.1415...

Answer 1

$$\lim_{\theta \to 0} -\frac{\theta\pi}{90} = \lim_{\theta \to 0} \sin(2\theta) $$

The above is correct, but it does not imply the following line:

$$\pi = -\lim_{\theta \to 0} \frac{90\sin(2\theta)}{\theta}$$

The fallacy here is expecting (or handwaving) that $\displaystyle \,\lim_{\theta \to 0} f(\theta) = \lim_{\theta \to 0} g(\theta) \implies \lim_{\theta \to 0} \frac{f(\theta)}{g(\theta)}=1\,$, but this latter implication does not necessarily hold true when both limits are $\,0\,$.

Compare for example to $\displaystyle\,\lim_{\theta \to 0} \theta + \lim_{\theta \to 0} \theta = 0 \implies \lim_{\theta \to 0} \theta = -\lim_{\theta \to 0} \theta \implies 1 = -\lim_{\theta \to 0} \dfrac{\theta}{\theta}=-1\,$.

Answer 2

Your mathematics is good up to

$$\lim_{\theta \to 0} -\frac{\theta\pi}{90} = \lim_{\theta \to 0} \sin(2\theta)$$ which is simply $$0=0$$ which is the same as $$-0=0$$

You want to manipulate $$0=0$$ by dividing both side by $0$ to generate the undefined $$0/0$$ and get a negative value for $\pi $

Well if notice $$\lim_{\theta \to 0} -\frac{\theta\pi}{90}= \lim_{\theta \to 0} \frac{\theta\pi}{90}$$ and with the same argument you can get $\pi = 3.14..$