The Limits of Geometry

Geometry Level pending

Find the limit as n approaches infinity. You can assume that anything inside the sine functions will be evaluated as degrees.

Please give a precision of 6 digits after the decimal place in your answer.

l i m n n ( sin ( 1 2 ( 180 360 n ) ) sin ( 90 ° ) sin ( 1 2 ( 360 n ) ) sin ( 90 ° ) ) lim\quad n\longrightarrow \infty \quad \quad \quad n\left( \frac { \sin { \left( \frac { 1 }{ 2 } \left( 180-\frac { 360 }{ n } \right) \right) } }{ \sin { (90°) } } \ast \frac { \sin { \left( \frac { 1 }{ 2 } \left( \frac { 360 }{ n } \right) \right) } }{ \sin { \left( 90° \right) } } \right)


The answer is 3.141592.

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