Get the limit from <munder> <mo movablelimits="true" form="prefix">lim <mrow class="MJX-

Amber Quinn 2022-06-26 Answered
Get the limit from
lim x ( x π 2 arctan ( x ) sin ( t 2 ) d t )
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Answers (1)

hofyonlines5
Answered 2022-06-27 Author has 12 answers
You have
lim x x π / 2 arctan x sin ( t 2 ) d t = lim x π / 2 arctan x sin ( t 2 ) d t 1 x = lim x 1 1 + x 2 sin ( arctan 2 x ) 1 x 2 = lim x x 2 1 + x 2 sin ( arctan 2 x ) = sin ( π 2 4 ) .
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