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Question Number 114239 by bemath last updated on 18/Sep/20
limx→∞cos(4x)3−1cos(2x)−cos(4x)?
Commented by bemath last updated on 18/Sep/20
limx→∞1−8x23−1(1−2x2)−(1−8x2)=limx→∞(1−83x2)−1(6x2)=−83.6=−49
Answered by Dwaipayan Shikari last updated on 18/Sep/20
limx→∞1−12(4x)23−12sin3xsin2x=1−83x2−12.6x2=−49
Answered by mathmax by abdo last updated on 19/Sep/20
letf(x)=(cos(4x))13−1cos(2x)−cos(4x)herewedothechangement1x=t⇒f(x)=f(1t)=(cos(4t))13−1cos(2t)−cos(4t)(x→∞⇒t→0)⇒cos(4t)∼1−8t2,cos(2t)∼1−2t2⇒f(1t)∼(1−8t2)13−11−2t2−1+8t2∼1−83t2−16t2=−818=−49⇒limx→+∞f(x)=−49
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