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Question Number 123874 by john_santu last updated on 29/Nov/20

 (1)lim_(θ→0)  ((θ−sin θcos θ)/(tan θ−θ))   (2) lim_(x→0)  ((sin 3x−3x+x^2 )/(sin xsin 2x))

(1)limθ0θsinθcosθtanθθ(2)limx0sin3x3x+x2sinxsin2x

Answered by liberty last updated on 29/Nov/20

 solve this question   (1) lim_(θ→0)  ((θ−sin θcos θ)/(tan θ−θ))  (2) lim_(x→0)  ((sin 3x−3x+x^2 )/(sin x sin 2x))  Solution :   (1) by L′Hopital rule    lim_(θ→0)  ((θ−(1/2)sin 2θ)/(tan θ−θ)) = lim_(θ→0)  ((1−cos 2θ)/(sec^2 θ−1))   lim_(θ→0)  ((cos^2 θ(2sin^2 ((θ/2))))/(sin^2 θ)) = lim_(θ→0)  2(((sin (θ/2))/(tan θ)))^2 = (1/2)  (2) lim_(x→0)  ((sin 3x−3x+x^2 )/(−(1/2)(cos 3x−cos x))) =    −2×lim_(x→0)  ((3cos 3x−3+2x)/(−3sin 3x+sin x)) = −2×lim_(x→0)  ((−9sin 3x+2)/(−9cos 3x+cos x))  = −2× [(2/(−8)) ] = (1/2).

solvethisquestion(1)limθ0θsinθcosθtanθθ(2)limx0sin3x3x+x2sinxsin2xSolution:(1)byLHopitalrulelimθ0θ12sin2θtanθθ=limθ01cos2θsec2θ1limθ0cos2θ(2sin2(θ2))sin2θ=limθ02(sinθ2tanθ)2=12(2)limx0sin3x3x+x212(cos3xcosx)=2×limx03cos3x3+2x3sin3x+sinx=2×limx09sin3x+29cos3x+cosx=2×[28]=12.

Answered by Dwaipayan Shikari last updated on 29/Nov/20

1)lim_(x→0) ((x−(1/2)sin2x)/(x+(x^3 /3)−x))=((x−x+(8/(12))x^3 )/(x^3 /3))=2  (sinx=x−(x^3 /6))  2)lim_(x→0) ((3sinx−4sin^3 x−3x+x^2 )/(x.2x))=((3x−4x^3 −3x+x^2 )/(2x^2 ))=(1/2)−2x  =(1/2)    (sinx→x)

1)limx0x12sin2xx+x33x=xx+812x3x33=2(sinx=xx36)2)limx03sinx4sin3x3x+x2x.2x=3x4x33x+x22x2=122x=12(sinxx)

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