Where is the mistake in this solution of lim x → 1 1 − x 2 sin ⁡ ( π x ) ?The answer ought to be 2 π , but I end up with 0: lim x → 1 1 − x 2 sin ⁡ ( π x ) = lim y → 0 1 − ( y + 1 ) 2 sin ⁡ ( π ( y + 1 ) ) = lim y → 0 π ( y + 1 ) sin ⁡ ( π ( y + 1 ) ) 1 − ( y + 1 ) 2 π ( y + 1 ) = lim y → 0 1 − ( y + 1 ) 2 π ( y + 1 ) = 0Where and why is my solution incorrect?

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Where is the mistake in this solution of       lim          x      →      1                          1        −                  x          2                            sin        ⁡        (        π        x        )            ?The answer ought to be       2    π  , but I end up with 0:      lim          x      →      1                          1        −                  x          2                            sin        ⁡        (        π        x        )              =       lim          y      →      0                          1        −        (        y        +        1                  )          2                            sin        ⁡        (        π        (        y        +        1        )        )              =       lim          y      →      0                          π        (        y        +        1        )                    sin        ⁡        (        π        (        y        +        1        )        )                            1        −        (        y        +        1                  )          2                            π        (        y        +        1        )              =       lim          y      →      0                          1        −        (        y        +        1                  )          2                            π        (        y        +        1        )              =  0Where and why is my solution incorrect?

livin4him777lf · Accepted Answer

Your third equality attempts to make use of the rule       lim          x      →      0            x          sin      ⁡      x        =  1, but note that yours has   y  →  0 yet the argument is not y, it is   π  (  y  +  1  ), which does not go to zero. That&#039;s where your work goes wrong.

Boilanubjaini8f · Accepted Answer

lim          y      →      0                          π        (        y        +        1        )                    sin        ⁡        (        π        (        y        +        1        )        )              =      π    0    ≠  1

Where is the mistake in this solution of <munder> <mo movablelimits="true" form="prefix">lim

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