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Limits Continuity and Differentiability question

2019 · 12 Apr · Shift 2 · Q35
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  5. /2019 · 12 Apr · Shift 2 · Q35

Limits Continuity and Differentiability question

2019 · 12 Apr · Shift 2 · Q35

JEE MainMathematicsLimits Continuity and DifferentiabilityMCQ+4 / −1
lim⁡x→0x+2sin⁡xx2+2sin⁡x+1−sin⁡2x−x+1\mathop {\lim }\limits_{x \to 0} {{x + 2\sin x} \over {\sqrt {{x^2} + 2\sin x + 1} - \sqrt {{{\sin }^2}x - x + 1} }}x→0lim​x2+2sinx+1​−sin2x−x+1​x+2sinx​ is :
  1. A
    6
  2. B
    1
  3. C
    3
  4. D
    2
View written solutionFree

Correct answer: D

  1. We need to evaluate L=lim⁡x→0x+2sin⁡xx2+2sin⁡x+1−sin⁡2x−x+1.L=\lim_{x\to 0}\frac{x+2\sin x}{\sqrt{x^2+2\sin x+1}-\sqrt{\sin^2 x-x+1}}.L=limx→0​x2+2sinx+1​−sin2x−x+1​x+2sinx​.

  2. Since the denominator is a difference of square roots, rationalize it: L=lim⁡x→0x+2sin⁡xx2+2sin⁡x+1−sin⁡2x−x+1⋅x2+2sin⁡x+1+sin⁡2x−x+1x2+2sin⁡x+1+sin⁡2x−x+1.L=\lim_{x\to 0}\frac{x+2\sin x}{\sqrt{x^2+2\sin x+1}-\sqrt{\sin^2 x-x+1}}\cdot \frac{\sqrt{x^2+2\sin x+1}+\sqrt{\sin^2 x-x+1}}{\sqrt{x^2+2\sin x+1}+\sqrt{\sin^2 x-x+1}}.L=limx→0​x2+2sinx+1​−sin2x−x+1​x+2sinx​⋅x2+2sinx+1​+sin2x−x+1​x2+2sinx+1​+sin2x−x+1​​.

So, L=lim⁡x→0(x+2sin⁡x)(x2+2sin⁡x+1+sin⁡2x−x+1)(x2+2sin⁡x+1)−(sin⁡2x−x+1).L=\lim_{x\to 0}\frac{(x+2\sin x)\left(\sqrt{x^2+2\sin x+1}+\sqrt{\sin^2 x-x+1}\right)}{(x^2+2\sin x+1)-(\sin^2 x-x+1)}.L=limx→0​(x2+2sinx+1)−(sin2x−x+1)(x+2sinx)(x2+2sinx+1​+sin2x−x+1​)​.

  1. Simplify the denominator: (x2+2sin⁡x+1)−(sin⁡2x−x+1)=x2−sin⁡2x+x+2sin⁡x.(x^2+2\sin x+1)-(\sin^2 x-x+1)=x^2-\sin^2 x+x+2\sin x.(x2+2sinx+1)−(sin2x−x+1)=x2−sin2x+x+2sinx.

Thus, L=lim⁡x→0(x+2sin⁡x)(x2+2sin⁡x+1+sin⁡2x−x+1)x2−sin⁡2x+x+2sin⁡x.L=\lim_{x\to 0}\frac{(x+2\sin x)\left(\sqrt{x^2+2\sin x+1}+\sqrt{\sin^2 x-x+1}\right)}{x^2-\sin^2 x+x+2\sin x}.L=limx→0​x2−sin2x+x+2sinx(x+2sinx)(x2+2sinx+1​+sin2x−x+1​)​.

  1. Factor the denominator: x2−sin⁡2x=(x−sin⁡x)(x+sin⁡x),x^2-\sin^2 x=(x-\sin x)(x+\sin x),x2−sin2x=(x−sinx)(x+sinx), but the more useful observation is x2−sin⁡2x+x+2sin⁡x=(x2−sin⁡2x)+(x+2sin⁡x).x^2-\sin^2 x+x+2\sin x=(x^2-\sin^2 x)+(x+2\sin x).x2−sin2x+x+2sinx=(x2−sin2x)+(x+2sinx). As x→0x\to 0x→0, we know x2−sin⁡2x→0x^2-\sin^2 x\to 0x2−sin2x→0 of higher order, while x+2sin⁡x∼3xx+2\sin x\sim 3xx+2sinx∼3x.

So near x=0x=0x=0, x2−sin⁡2x+x+2sin⁡x∼x+2sin⁡x.x^2-\sin^2 x+x+2\sin x \sim x+2\sin x.x2−sin2x+x+2sinx∼x+2sinx. Hence, L=lim⁡x→0(x2+2sin⁡x+1+sin⁡2x−x+1).L=\lim_{x\to 0}\left(\sqrt{x^2+2\sin x+1}+\sqrt{\sin^2 x-x+1}\right).L=limx→0​(x2+2sinx+1​+sin2x−x+1​).

  1. Now substitute x=0x=0x=0: 0+0+1+0−0+1=1+1=2.\sqrt{0+0+1}+\sqrt{0-0+1}=1+1=2.0+0+1​+0−0+1​=1+1=2.

Therefore, L=2.L=2.L=2.

  1. Checking options:
  • A: 666 ✗
  • B: 111 ✗
  • C: 333 ✗
  • D: 222 ✓

So the correct option is D.

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