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Question Number 111907 by bemath last updated on 05/Sep/20
limx→π201−tan5xsin5x−cos5x?
Answered by john santu last updated on 05/Sep/20
anotherwaylimx→π20cos5x−sin5xsin5x−cos5x.1cos5x=−limx→π201cos5x=−1cos(π4)=−2
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