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Question Number 107414 by bemath last updated on 10/Aug/20
⊚BeMath⊚limx→02−3cos6xcos42xcos24x+cosx36x2
Answered by ajfour last updated on 10/Aug/20
L=limx→02−3(1−sin2x)3(1−sin22x)2(1−sin24x)+cosx36x2=limx→02−3(1−3sin2x)(1−2sin22x)(1−sin24x)+cosx36x2=limsinx=h→02−3(1−3h2){1−8h2(1−h2)}{1−4[4(h2)(1−h2)](1−2h2)2}+1−h236h2=limh2=k→02−3{1−3k){1−8k(1−k){1−16k+...}+1−k36k=limk→02−3(1−3k)(1−8k+..)(1−16k+..)+1−k36k=limk→02−3(1−27k+...)+1−k236k=81−1236=16172.
Answered by john santu last updated on 11/Aug/20
⋇JS⋇limx→02−3(1−x22)6(1−4x22)4(1−16x22)2+1−x2236x2=limx→03−3(1−3x2)(1−8x2)(1−16x2)−x2236x2=setx2=tlimt→03−3(1−3t)(1−8t)(1−16t)−t236t=136limt→03−3(1−11t+24t2)(1−16t)−t2t=136limt→03−3(1−27t+o(t))−t2t=136limt→081t−t2+3o(t)t=136×(1612)=16172
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