Finding $\lim_{x \to 0} \frac{180\sin x}{x}$ I am in ninth grade so I am an amateur in mathematics and with no training in limits. I self derived this limit to find the value of $\pi$.
I imagined a circle to be composed of infinitely small right triangles.
The central angle is $x$ and hence the side closest to the circum will have side $r \sin x $ (where $r$ is the radius) 
There will be $360/x$ such triangles. 
Comparing this to the already existing circumference formula we will get 
$$\pi = \lim_{x \to 0}  \frac{180\sin x}{x}$$
But I have seen in Math Stack Exchange 
itself that:
$$\lim_{x \to 0}\frac{\sin x}{x}=1 $$
As seen in the first line of the accepted answer to this question here:
How to find $\lim_{x\to 0} \frac{1-\cos x \sqrt{\cos 2x}}{x^2}$
But shouldn't it be equal to $\pi/180$ as per my derivation? Where is my mistake?
 A: Make sure you are properly differentiating between degrees vs radians. As far as I know, the answer is $\frac{\pi}{180}$ if you are in degrees, and $1$ if you are in radians. 
A: From your calculations we get the side length of each right-triangle opposite the angle $x^{\circ}$ where the hypotenuse is of length $r$ is given by:$$r\times\sin(x^{\circ})$$
And the number of such triangles is given by:$$\frac{360^{\circ}}{x^{\circ}}$$
Therefore if we add up all these side lengths we get an approximation to the circumference of a circle of radius $r$ as:$$C\approx\frac{360^{\circ}}{x^{\circ}}\times r\times\sin(x^{\circ})$$
If we take this to the limit as $x\to0$ we obtain:$$C=\lim_{x\to0}\frac{360^{\circ}}{x^{\circ}}\times r\times\sin(x^{\circ})=360^{\circ}r\lim_{x\to0}\frac{\sin(x^{\circ})}{x^{\circ}}=360^{\circ}r$$
We also know the the circumference of circle of radius $r$ is given by $2\pi r$. We can therefore conclude that:$$\require{cancel}2\pi\cancel{r}=360^{\circ}\cancel{r}$$$$\therefore \pi=180^{\circ}$$
This gives us the conversion of radians to degrees.
