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JEE Main 2015 · 4 Apr · Shift 1, Mathematics question 70

lim⁡x→0(1−cos⁡2x)(3+cos⁡x)xtan⁡4x\displaystyle\lim_{x\to0}\frac{(1-\cos2x)(3+\cos x)}{x\tan4x} is equal to :

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Answer and worked solution

Official answer: C

1−cos⁡2x=2sin⁡2x≈2x21-\cos2x=2\sin^2x\approx2x^2, 3+cos⁡x→43+\cos x\to4 and xtan⁡4x≈4x2x\tan4x\approx4x^2:
lim⁡=2x2×44x2=2.\lim=\frac{2x^2\times4}{4x^2}=2.

Topics:Limits
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