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Question Number 72071 by aliesam last updated on 23/Oct/19
a)limx→03x−sin(2x)2x−sin(3x)
Commented by mathmax by abdo last updated on 24/Oct/19
sin(2x)∼2xandsin(3x)∼3x(x→0)⇒3x−sin(2x)2x−sin(3x)∼3x−2x2x−3x∼−1⇒limx→03x−sin(2x)2x−sin(3x)=−1
Answered by Tanmay chaudhury last updated on 23/Oct/19
limx→03−(sin2x2x)×22−(sin3x3x)×3=3−1×22−1×3=−1
Answered by malwaan last updated on 24/Oct/19
limx→03x−sin(2x)2x−sin(3x)=3−22−3=−1
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