Why does lim x → π sin ⁡ ( m x ) sin ⁡ ( n x ) = ( − 1 ) m − n m n , for positive naturals m and n?

Question

Why does       lim          x      →      π                  sin      ⁡      (      m      x      )              sin      ⁡      (      n      x      )        =            (      −      1      )              m      −      n              m    n  , for positive naturals m and n?

Hailee Henderson · Accepted Answer

You can substitute   x  =  y  +  π, so you get      lim          y      →      0                  sin      ⁡      (      m      y      +      m      π      )              sin      ⁡      (      n      y      +      n      π      )      Since   sin  ⁡  (  α  +  k  π  )  =  (  −  1      )    k    sin  ⁡  α, you get      lim          y      →      0                  (      −      1              )        m            sin      ⁡      (      m      y      )              m      y                  n      y              (      −      1              )        n            sin      ⁡      (      n      y      )            m    n  and you can apply the basic limit      lim          x      →      0                  sin      ⁡      x        x    =  1so you get  (  −  1      )    m    (  −  1      )          −      n            m    n    =  (  −  1      )          m      −      n            m    n

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