π Archimedes Method

Can also look at this this way:

 

When \(\theta << 1 \) then \(\sin \theta \rightarrow \theta\)

  

As \(x \rightarrow \infty \quad \frac{a}{x}<<1\)  so  \(x \times \sin \frac{a}{x} \rightarrow x\times \frac{a}{x} \rightarrow a\)

MWizzard,

 

I have been playing with it in Desmos and with Wolfram Alpha.

In Desmos for some strange reason, the left hand side th angle is taken to be in degrees and the answer is is given in radians, I do not understand why it has done this.

 

Generally speaking,    

 

\(\displaystyle \lim_{x \rightarrow \infty} \;xsin(\frac{a}{x})=a\)

 

If you know calculus I think this can be shown with L'Hopital's rule.  

I have no intuitive understanding of why this is so but if you know L'Hopital's rule I can show your that it is true.

 

\(\displaystyle \lim_{x \rightarrow \infty} \;xsin(\frac{a}{x})=a\\ =\displaystyle \lim_{x \rightarrow \infty} \;\frac{sin(\frac{a}{x})}{\frac{1}{x}}=\frac{0}{0}\\ \mbox{This is indeterminant and we can use L'Hopital's rule to find the answer}\\ =\displaystyle \lim_{x \rightarrow \infty} \;\frac{\frac{d}{dx}sin(\frac{a}{x})}{\frac{d}{dx}\frac{1}{x}}\\ =\displaystyle \lim_{x \rightarrow \infty} \;\frac{\frac{d}{dx}sin(ax^{-1})}{\frac{d}{dx}x^{-1}}\\ =\displaystyle \lim_{x \rightarrow \infty} \;\frac{(-ax^{-2}cos(ax^{-1}))}{-1x^{-2}}\\ =\displaystyle \lim_{x \rightarrow \infty} \;\;+acos(ax^{-1})\\ =+acos(0)\\ =\;a \)

It is very well explained at this link:

http://betterexplained.com/articles/prehistoric-calculus-discovering-pi/