NTA Abhyas JEE Main2020MathematicsLimitsPractice
If l i m x → 0 sin 2 x - a sin x x 3 exists finitely, then the value of a is
Options
- A0
- B2
- C1
- D4
Correct answer
B. 2
Step-by-step solution
l i m x → 0 sin ⁡ x 2 cos ⁡ x - a x ⋅ x 2 = l i m x → 0 sin ⁡ x x 2 cos ⁡ x - a x 2 For this limit to exists finitely, l i m x → 0 2 cos ⁡ x - a x 2 = f i n i t e ∴ It must be 0 0 form ∴   2 cos ⁡ 0 - a = 0 ⇒ a = 2