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Question Number 107414 by bemath last updated on 10/Aug/20

                   ⊚BeMath⊚  lim_(x→0)  ((2−3cos^6 x cos^4 2x cos^2 4x+cos x)/(36x^2 ))

BeMathlimx023cos6xcos42xcos24x+cosx36x2

Answered by ajfour last updated on 10/Aug/20

L=lim_(x→0) ((2−3(1−sin^2 x)^3 (1−sin^2 2x)^2 (1−sin^2 4x)+cos x)/(36x^2 ))    = lim_(x→0) ((2−3(1−3sin^2 x)(1−2sin^2 2x)(1−sin^2 4x)+cos x)/(36x^2 ))    = lim_(sin x=h→0) ((2−3(1−3h^2 ){1−8h^2 (1−h^2 )}{1−4[4(h^2 )(1−h^2 )](1−2h^2 )^2 }+(√(1−h^2 )))/(36h^2 ))    =lim_(h^2 =k→0) ((2−3{1−3k){1−8k(1−k){1−16k+...}+(√(1−k)))/(36k))    = lim_(k→0) ((2−3(1−3k)(1−8k+..)(1−16k+..)+(√(1−k)))/(36k))    = lim_(k→0) ((2−3(1−27k+...)+1−(k/2))/(36k))    = ((81−(1/2))/(36)) = ((161)/(72)) .

L=limx023(1sin2x)3(1sin22x)2(1sin24x)+cosx36x2=limx023(13sin2x)(12sin22x)(1sin24x)+cosx36x2=limsinx=h023(13h2){18h2(1h2)}{14[4(h2)(1h2)](12h2)2}+1h236h2=limh2=k023{13k){18k(1k){116k+...}+1k36k=limk023(13k)(18k+..)(116k+..)+1k36k=limk023(127k+...)+1k236k=811236=16172.

Answered by john santu last updated on 11/Aug/20

               ⋇JS⋇  lim_(x→0) ((2−3(1−(x^2 /2))^6 (1−((4x^2 )/2))^4 (1−((16x^2 )/2))^2 +1−(x^2 /2))/(36x^2 ))=  lim_(x→0) ((3−3(1−3x^2 )(1−8x^2 )(1−16x^2 )−(x^2 /2))/(36x^2 ))=  set x^2  = t  lim_(t→0)  ((3−3(1−3t)(1−8t)(1−16t)−(t/2))/(36t))=  (1/(36)) lim_(t→0)  ((3−3(1−11t+24t^2 )(1−16t)−(t/2))/t)=  (1/(36)) lim_(t→0)  ((3−3(1−27t+o(t))−(t/2))/t) =  (1/(36)) lim_(t→0)  ((81t−(t/2)+3 o(t))/t)=(1/(36))×(((161)/2))  = ((161)/(72))

JSlimx023(1x22)6(14x22)4(116x22)2+1x2236x2=limx033(13x2)(18x2)(116x2)x2236x2=setx2=tlimt033(13t)(18t)(116t)t236t=136limt033(111t+24t2)(116t)t2t=136limt033(127t+o(t))t2t=136limt081tt2+3o(t)t=136×(1612)=16172

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