NTA Abhyas JEE Main2020MathematicsLimitsPractice
If lim x → 0 sin 2 x - a sin x x 3 3 = L exists finitely, then the absolute value of L is equal to
Correct answer
27
Step-by-step solution
L = l i m x → 0 27 sin ⁡ x 2 cos ⁡ x - a x ⋅ x 2 = l i m x → 0 sin ⁡ x x 2 cos ⁡ x - a x 2 × 27 This limit to exists finitely lim x → 0 ⁡ 2 cos ⁡ x - a x 2 = finite ∴ It must be 0 0 form ∴ 2 cos ⁡ 0 - a = 0 ⇒ a = 2 ⇒ L = lim x → 0 ⁡ sin ⁡ x x × 2 cos ⁡ x - 1 x 2 × 27 ⇒ L = 1 × 2 × - 1 2 × 27 = - 27