Solving a trig limit without L'Hopital: lim x → 0 tan 2 ⁡ ( π x ) 2 ( π x ) 2

Question

Solving a trig limit without L&#039;Hopital:       lim          x      →      0                          tan        2            ⁡      (      π      x      )              2      (      π      x              )        2

Aiden Norman · Accepted Answer

If one knows that      lim          x      →      0                  sin      ⁡      x        x    =  1then one may write, as   x  →  0,                    tan        2            ⁡      (      π      x      )              2              π        2                    x        2              =      1    2    ⋅            (                        sin          ⁡          (          π          x          )                          π          x                    )        2    ⋅      1                  cos        2            ⁡      (      π      x      )      and conclude easily since             1                        cos          2                ⁡        (        π        x        )              →    1   as   x  →  0

Solving a trig limit without L'Hopital: <munder> <mo movablelimits="true" form="prefix">lim

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