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Question Number 84492 by john santu last updated on 13/Mar/20
limx→π3sin(x−π3)1−2cos(x)=
Commented by john santu last updated on 13/Mar/20
limx→0sin(x)1−2cos(x+π3)=limx→0cos(x)2sin(x+π3)=12×32=13
otherwaysin(x−π3)≈(x−π3)cos(x)≈12−32(x−π3)limx→π3x−π31−2[12−32(x−π3)]=limx→π3(x−π3)3(x−π3)=13.
Commented by mathmax by abdo last updated on 13/Mar/20
letl(x)=sin(x−π3)1−2cosxchangementx−π3=tgivel(x)=A(t)=sint1−2cos(t+π3)=sint1−2(costcosπ3−sintsin(π3))=sint1−2(12cost−32sint)=sint1−cost+3sintx→π3⇔t→0andA(t)∼tt22+3t=1t2+3⇒limt→0A(t)=13⇒limx→0l(x)=13
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