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Limits Continuity and Differentiability question

2002 · Shift 0 · Q64
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  5. /2002 · Shift 0 · Q64

Limits Continuity and Differentiability question

2002 · Shift 0 · Q64

JEE MainMathematicsLimits Continuity and DifferentiabilityMCQ+4 / −1
lim⁡x→01−cos⁡2x2x\mathop {\lim }\limits_{x \to 0} {{\sqrt {1 - \cos 2x} } \over {\sqrt 2 x}}x→0lim​2​x1−cos2x​​ is
  1. A
    111
  2. B
    −1-1−1
  3. C
    zero
  4. D
    does not exist
View written solutionFree

Correct answer: D

  1. We need to evaluate
lim⁡x→01−cos⁡2x2 x.\lim_{x\to 0}\frac{\sqrt{1-\cos 2x}}{\sqrt{2}\,x}.x→0lim​2​x1−cos2x​​.
  1. Use the identity
1−cos⁡2x=2sin⁡2x.1-\cos 2x = 2\sin^2 x.1−cos2x=2sin2x.

So,

1−cos⁡2x=2sin⁡2x=2 ∣sin⁡x∣.\sqrt{1-\cos 2x}=\sqrt{2\sin^2 x}=\sqrt{2}\,|\sin x|.1−cos2x​=2sin2x​=2​∣sinx∣.

Hence the limit becomes

lim⁡x→02 ∣sin⁡x∣2 x=lim⁡x→0∣sin⁡x∣x.\lim_{x\to 0}\frac{\sqrt{2}\,|\sin x|}{\sqrt{2}\,x} =\lim_{x\to 0}\frac{|\sin x|}{x}.x→0lim​2​x2​∣sinx∣​=x→0lim​x∣sinx∣​.
  1. Now examine right-hand and left-hand limits.
  • As x→0+x\to 0^+x→0+, sin⁡x>0\sin x>0sinx>0, so ∣sin⁡x∣=sin⁡x|\sin x|=\sin x∣sinx∣=sinx. Thus,
lim⁡x→0+∣sin⁡x∣x=lim⁡x→0+sin⁡xx=1.\lim_{x\to 0^+}\frac{|\sin x|}{x} =\lim_{x\to 0^+}\frac{\sin x}{x}=1.x→0+lim​x∣sinx∣​=x→0+lim​xsinx​=1.
  • As x→0−x\to 0^-x→0−, sin⁡x<0\sin x<0sinx<0, so ∣sin⁡x∣=−sin⁡x|\sin x|=-\sin x∣sinx∣=−sinx. Thus,
lim⁡x→0−∣sin⁡x∣x=lim⁡x→0−−sin⁡xx=−1.\lim_{x\to 0^-}\frac{|\sin x|}{x} =\lim_{x\to 0^-}\frac{-\sin x}{x}=-1.x→0−lim​x∣sinx∣​=x→0−lim​x−sinx​=−1.

  1. Since the right-hand limit is 111 and the left-hand limit is −1-1−1, they are not equal. Therefore, the two-sided limit does not exist.

  2. Checking options:

  • A: 111 ❌
  • B: −1-1−1 ❌
  • C: zero ❌
  • D: does not exist ✅

Therefore, the correct answer is D.

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