Solving this trig limit without L'hopital: lim x → 1 sin ⁡ ( π x ) x − 1 One idea I had was to multiply both sides of the fractions by π xlike this lim x → 1 sin ⁡ ( π x ) π x π x 2 − π x to cancel out the sine but then I realized that I could not do that because x tends to 1, and not to 0. What can I do?

Question

Solving this trig limit without L&#039;hopital:       lim          x      →      1                  sin      ⁡      (      π      x      )              x      −      1       One idea I had was to multiply both sides of the fractions by  π  xlike this      lim          x      →      1                  sin      ⁡      (      π      x      )      π      x              π              x        2            −      π      x      to cancel out the sine but then I realized that I could not do that because x tends to 1, and not to 0. What can I do?

Madden Luna · Accepted Answer

lim          h      →      0                  sin      ⁡      (      π      (      1      −      h      )      )              1      −      1      −      h                          (Multiply and divide by pi)                              lim                      h            →            0                                                sin            ⁡            (            π            h            )            π                                −            π            h                                −  π

Sylvia Byrd · Accepted Answer

lim                      x            →            1                                                sin            ⁡            (            π            x            )                                1            −            x                                              =                              lim                      h            →            0                                                sin            ⁡            (            π            (            1            −            h            )            )                    h                                                  =                              lim                      h            →            0                                                sin            ⁡            (            −            π            h            )                    h                                    (common limit)                                  =                              lim                      h            →            0                                                −            sin            ⁡            (            π            h            )            π                                π            h                                                            =                    −        π        .

Solving this trig limit without L'hopital: <munder> <mo movablelimits="true" form="prefix">

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