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Limits Continuity and Differentiability question

2011 · Shift 0 · Q31
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  5. /2011 · Shift 0 · Q31

Limits Continuity and Differentiability question

2011 · Shift 0 · Q31

JEE MainMathematicsLimits Continuity and DifferentiabilityMCQ+4 / −1
lim⁡x→2(1−cos⁡{2(x−2)}x−2)\mathop {\lim }\limits_{x \to 2} \left( {{{\sqrt {1 - \cos \left\{ {2(x - 2)} \right\}} } \over {x - 2}}} \right)x→2lim​(x−21−cos{2(x−2)}​​)
  1. A
    Equals 2\sqrt 22​
  2. B
    Equals −2-\sqrt 2−2​
  3. C
    Equals 12{1 \over {\sqrt 2 }}2​1​
  4. D
    does not exist
View written solutionFree

Correct answer: D

  1. Let h=x−2.h=x-2.h=x−2. Then as x→2x\to 2x→2, we have h→0h\to 0h→0.

The limit becomes lim⁡h→01−cos⁡(2h)h.\lim_{h\to 0} \frac{\sqrt{1-\cos(2h)}}{h}.limh→0​h1−cos(2h)​​.

  1. Use the identity 1−cos⁡(2h)=2sin⁡2h.1-\cos(2h)=2\sin^2 h.1−cos(2h)=2sin2h. So, 1−cos⁡(2h)=2sin⁡2h=2 ∣sin⁡h∣.\sqrt{1-\cos(2h)}=\sqrt{2\sin^2 h}=\sqrt{2}\,|\sin h|.1−cos(2h)​=2sin2h​=2​∣sinh∣.

Hence, 1−cos⁡(2h)h=2 ∣sin⁡h∣h.\frac{\sqrt{1-\cos(2h)}}{h}=\sqrt{2}\,\frac{|\sin h|}{h}.h1−cos(2h)​​=2​h∣sinh∣​.

  1. Now examine the one-sided limits.

  • If h→0+h\to 0^+h→0+, then sin⁡h>0\sin h>0sinh>0 for small hhh, so ∣sin⁡h∣=sin⁡h|\sin h|=\sin h∣sinh∣=sinh. Thus 2 ∣sin⁡h∣h=2 sin⁡hh→2.\sqrt{2}\,\frac{|\sin h|}{h}=\sqrt{2}\,\frac{\sin h}{h}\to \sqrt{2}.2​h∣sinh∣​=2​hsinh​→2​.

  • If h→0−h\to 0^-h→0−, then sin⁡h<0\sin h<0sinh<0 for small hhh, so ∣sin⁡h∣=−sin⁡h|\sin h|=-\sin h∣sinh∣=−sinh. Thus 2 ∣sin⁡h∣h=2 −sin⁡hh→−2 lim⁡h→0sin⁡hh=−2.\sqrt{2}\,\frac{|\sin h|}{h}=\sqrt{2}\,\frac{-\sin h}{h}\to -\sqrt{2}\,\lim_{h\to 0}\frac{\sin h}{h}=-\sqrt{2}.2​h∣sinh∣​=2​h−sinh​→−2​limh→0​hsinh​=−2​.

  1. Since the left-hand and right-hand limits are different, lim⁡x→21−cos⁡(2(x−2))x−2\lim_{x\to 2} \frac{\sqrt{1-\cos(2(x-2))}}{x-2}limx→2​x−21−cos(2(x−2))​​ does not exist.

  2. Therefore, the correct option is: D: does not exist\boxed{\text{D: does not exist}}D: does not exist​

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