ExamHabit
Practise free

1999

JEE Main 1999 Mathematics question on Limits Continuity and Differentiability

lim⁡x→0xtan⁡2x−2xtan⁡x(1−cos⁡2x)2\lim_{x \to 0} \dfrac{x \tan 2x - 2x \tan x}{(1 - \cos 2x)^2} is
  1. A.
    2
  2. B.
    −2-2
  3. C.
    12\dfrac{1}{2}
  4. D.
    −12-\dfrac{1}{2}
Show answer and solution

Correct option: C

Solution

Use the small-angle expansions: tan x ≈ x + x³/3 + ..., cos 2x ≈ 1 - 2x² + ..., so 1 - cos 2x ≈ 2x².

Numerator: x·tan2x - 2x·tanx
= x(2x + 8x³/3 + ...) - 2x(x + x³/3 + ...)
= (2x² + 8x⁴/3 + ...) - (2x² + 2x⁴/3 + ...)
= (8/3 - 2/3)x⁴ + ... = 2x⁴.

Denominator: (1 - cos2x)² ≈ (2x²)² = 4x⁴.

Limit = 2x⁴ / 4x⁴ = 1/2.

[1999]

Practise Limits Continuity and Differentiability the way it is asked

A timed set of real JEE Main Limits Continuity and Differentiability questions, graded as the exam grades. No sign-up.

Start a Limits Continuity and Differentiability test

More: all JEE Main Limits Continuity and Differentiability questions · JEE Main 1999 · Mathematics chapter list