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Question Number 184796 by Spillover last updated on 11/Jan/23
Evaluatelimx→π63sinx−cosxx−π6
Commented by MJS_new last updated on 11/Jan/23
2
thesearealleasywithl′Hopital^
Answered by aba last updated on 12/Jan/23
lett=x−π6⇒x=t+π6limx→π63sin(x)−cos(x)x−π6=limt→03sin(t+π6)−cos(t+π6)t=limt→03(sin(t)cos(π6)+cos(t)sin(π6))−(cos(t)cos(π6)−sin(t)sin(π6))t=limt→03(3sin(t)+cos(t))−(3cos(t)−sin(t))2t=limt→03sin(t)+sin(t)+3sin(t)−3cos(t)2t=limt→04sin(t)2t=2
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