If I were to convert a quarter circle into isosceles triangles, I would find I have a triangle with 2 sides of length r (radius), and the other length is to be found, I'll call it "a".
If I split the quarter circle (90 degrees) into n segments, I would find the angle opposite to side "a" would be equal to $\frac{90}{n}$. Given the triangle is isosceles, the two angles opposite the sides of length r would be $\frac{1}{2}\times (180-\frac{90}{n})$, which simplifies to ($90 - \frac{45}{n}$).
Hence, the sin rule means;
$\frac{a}{sin(\frac{90}{n})}=\frac{r}{sin(90-\frac{45}{n})}$
This is the same as;
$a = \frac{r \times sin(\frac{90}{n})}{sin(90-\frac{45}{n})}$
If I wanted to find the sum of these sides, I would simply multiply it by n, given there are n sides, so;
$a = n\times\frac{ r \times sin(\frac{90}{n})}{sin(90-\frac{45}{n})}$
I could use this to calculate $\frac{1}{4}$ of the circumference of a circle. Given a circle has infinite sides, n should be infinite. Therefore;
As $\lim_{n \rightarrow \infty}(\frac{90}{n})=0$, and;
$\lim_{n \rightarrow \infty}(\frac{45}{n})=0$, then;
$a = n \times \frac{r \times sin(0)}{sin(90-0)}$, so;
$a = n \times \frac{r \times 0}{1}$, which means;
$a = n \times 0$, which means that because any number times 0 is 0, a must be equal to 0. Hence, as this is a quarter of the circumference, the circumference must be equal to $4 \times 0 = 0$.
As $circumference = \pi \times d$, then $\pi = \frac{circumference}{d}$. As circumference is 0, $\pi = \frac{0}{d}$, so $\pi = 0$.
Of course, it defies logic for circumference to always be equal to 0, and therefore it similarly defies logic for pi to be equal to 0, so where have I gone wrong in getting to this answer?
Your work in computing the limiting value of $a$ is basically the same as the following, more obvious mistake:
$$ 1 = \lim_{n \to \infty} 1 = \lim_{n \to \infty}\left( n \cdot \frac{1}{n} \right) \mathrel{\color{red}=} n \cdot \lim_{n \to \infty} \frac{1}{n} = n \cdot 0 = 0 $$