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Question Number 73330 by mathmax by abdo last updated on 10/Nov/19
findlimx→0ln(2−cos(2x))ln(1+xsin(3x))
Commented by mathmax by abdo last updated on 11/Nov/19
letf(x)=ln(2−cos(2x))ln(1+xsin(3x))wehavecos(2x)∼1−(2x)22=1−2x2⇒2−cos(2x)∼2−1+2x2=1+2x2andln(1+2x2)∼2x2(x∼0)⇒alsoxsin(3x)∼x(3x)=3x2⇒f(x)∼2x23x2⇒limx→0f(x)=23
Answered by Smail last updated on 10/Nov/19
ln(2−cos(2x))=ln(1+2sin2(x))∼02sin2(x)ln(1+xsin(3x))∼0xsin(3x)limx→0ln(2−cos(2x))ln(1+xsin(3x))=limx→02sin2xxsin(3x)=limx→02sinxx×sinxx×3xsin(3x)×13=23
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