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JEE Main 2019 · 8 April · Shift I

JEE Main 2019 Mathematics question on Limits Continuity and Differentiability

lim⁡x→0sin⁡2x2−1+cos⁡x\lim_{x \to 0} \dfrac{\sin^2 x}{\sqrt{2} - \sqrt{1 + \cos x}} equals
  1. A.
    424\sqrt{2}
  2. B.
    2\sqrt{2}
  3. C.
    222\sqrt{2}
  4. D.
    44
Show answer and solution

Correct option: A

Solution

Multiply numerator and denominator by (√2 + √(1+cosx)):

Denominator becomes: 2 - (1+cosx) = 1 - cosx

So the limit becomes: lim(x→0) [sin²x · (√2 + √(1+cosx))] / (1 - cosx)

Since sin²x = 1 - cos²x = (1-cosx)(1+cosx), this simplifies to:

lim(x→0) [(1-cosx)(1+cosx) · (√2 + √(1+cosx))] / (1-cosx)

= lim(x→0) (1+cosx)(√2 + √(1+cosx))

At x=0: cosx=1, so = (1+1)(√2 + √2) = 2 · 2√2 = 4√2

[JEE Main 2019 · 8 April · Shift I]

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