The day before PI

Here we know that $lim_{n \rightarrow \infty} \ 2n\ sin(\frac{90^{\circ}}{n})$

So now LaTeX: $lim_{n \rightarrow \infty} \ 2\dfrac{\ sin(\frac{90^{\circ}}{n})}{\dfrac{1}{n}}$

Lets take $\dfrac{1}{n} = m$

So $lim_{m \rightarrow 0} \ 2\dfrac{\ sin({90^{\circ}m})}{m}$

So Now $90^{\circ} = \dfrac{\pi}{2}$

So rewriting

$lim_{m \rightarrow 0} \ 2\dfrac{\ sin({\dfrac{\pi}{2}m})}{m}$

Now Multiplying $\dfrac{\pi}{2}$ to denominator and numerator we get,

$lim_{m \rightarrow 0} \ 2\dfrac{\pi}{2}\dfrac{\ sin({\dfrac{\pi}{2}m})}{\dfrac{\pi}{2}m}$

Now we know that $lim_{x \rightarrow 0} \dfrac{\sin x}{x}$ = $1$

So $lim_{m \rightarrow 0} \ 2\dfrac{\pi}{2}\dfrac{\ sin({\dfrac{\pi}{2}m})}{\dfrac{\pi}{2}m}$ = $2 . \dfrac{\pi}{2}$

So Limit = $\boxed{\pi}= \boxed{3.14}$ (Upto two decimal places)

Let's take a look at the first quadrant of the unit circle. Let's divide the angle of $90^{\circ}$ into n equals angles. If we connect the sides of these n corners, we will find n congruent triangles with 2 legs of length 1. We want to calculate the area of one triangle. To do that we need to find a base and a height. Let's call the base the opposite of the known angel of $\frac{90^{\circ}}{n}$ . Then with the sine rule we can calculate the length of the base: $base=\frac{sin(\frac{90^{\circ}}{n})}{sin(90^{\circ}-\frac{45^{\circ}}{2n}})$ . For the height we draw a line perpendicalur on the base. Now we can calculate the height by using the formula for the sinus: $sin(90^{\circ}-\frac{45^{\circ}}{2n}) = \frac{height}{1} = height$ . So the area of 1 triangle now is: $\frac{1}{2} \cdot sin(90^{\circ}-\frac{45^{\circ}}{2n})\cdot\frac{sin(\frac{90^{\circ}}{n})}{sin(90^{\circ}-\frac{45^{\circ}}{2n})} =\frac{1}{2}sin(\frac{90^{\circ}}{n})$ . Now since we have n triangles in the first quarter, we have a total area of $n \cdot \frac{1}{2}sin(\frac{90^{\circ}}{n})$ . Now it we take $lim_{n\rightarrow \infty}$ we find the total area of the first quadrant. From the formula for the area if this quadrant we now that this equals $\frac{\pi}{4}$ , so:

$lim_{n\rightarrow \infty} \frac{n}{2}sin(\frac{90^{\circ}}{n}) =\frac{\pi}{4}$ .

Now we know: $lim_{n\rightarrow \infty } 2n sin(\frac{90^{\circ}}{n}) = 4\cdot \frac{\pi}{4} = \pi$ .