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

In you answer, give a precision of 6 digits after the decimal place.

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

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