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Limits Continuity and Differentiability question

2015 · Shift 0 · Q25
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  5. /2015 · Shift 0 · Q25

Limits Continuity and Differentiability question

2015 · Shift 0 · Q25

JEE MainMathematicsLimits Continuity and DifferentiabilityMCQ+4 / −1
lim⁡x→0(1−cos⁡2x)(3+cos⁡x)xtan⁡4x\mathop {\lim }\limits_{x \to 0} {{\left( {1 - \cos 2x} \right)\left( {3 + \cos x} \right)} \over {x\tan 4x}}x→0lim​xtan4x(1−cos2x)(3+cosx)​ is equal to
  1. A
    2
  2. B
    12{1 \over 2}21​
  3. C
    4
  4. D
    3
View written solutionFree

Correct answer: A

  1. We need to evaluate
L=lim⁡x→0(1−cos⁡2x)(3+cos⁡x)xtan⁡4x.L=\lim_{x\to 0}\frac{(1-\cos 2x)(3+\cos x)}{x\tan 4x}.L=x→0lim​xtan4x(1−cos2x)(3+cosx)​.
  1. Use standard small-angle limits:
1−cos⁡2x=2sin⁡2x1-\cos 2x = 2\sin^2 x1−cos2x=2sin2x

and as x→0x\to 0x→0,

sin⁡x∼x,tan⁡4x∼4x,cos⁡x→1.\sin x \sim x, \qquad \tan 4x \sim 4x, \qquad \cos x \to 1.sinx∼x,tan4x∼4x,cosx→1.

So,

1−cos⁡2x=2sin⁡2x∼2x2,1-\cos 2x = 2\sin^2 x \sim 2x^2,1−cos2x=2sin2x∼2x2,

and

3+cos⁡x→3+1=4.3+\cos x \to 3+1=4.3+cosx→3+1=4.

Also,

xtan⁡4x∼x⋅4x=4x2.x\tan 4x \sim x\cdot 4x = 4x^2.xtan4x∼x⋅4x=4x2.
  1. Therefore,
L=lim⁡x→0(1−cos⁡2x)(3+cos⁡x)xtan⁡4x∼(2x2)(4)4x2=2.L=\lim_{x\to 0}\frac{(1-\cos 2x)(3+\cos x)}{x\tan 4x} \sim \frac{(2x^2)(4)}{4x^2}=2.L=x→0lim​xtan4x(1−cos2x)(3+cosx)​∼4x2(2x2)(4)​=2.
  1. Hence,
2\boxed{2}2​
  1. Checking options:
  • A: 222 ✅
  • B: 12\frac1221​ ❌
  • C: 444 ❌
  • D: 333 ❌

So the correct option is A.

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