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

Mohamed Mooney 2022-06-25 Answered
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 ) = 0
Where and why is my solution incorrect?
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Answers (2)

livin4him777lf
Answered 2022-06-26 Author has 14 answers
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's where your work goes wrong.

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Boilanubjaini8f
Answered 2022-06-27 Author has 6 answers
lim y 0 π ( y + 1 ) sin ( π ( y + 1 ) ) = π 0 1

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