JEE Main 2023 · 31 January · evening · Single correct · +4 / −1
lim(x → ∞) ((√(3 x+1)+√(3 x-1))⁶+(√(3 x+1)-√(3 x-1))⁶)/((x+√(x²-1))⁶+(x-√(x²-1))⁶) x³
- (a) is equal to 9
- (b) is equal to 27/2
- (c) does not exist
- (d) is equal to 27
Answer and solution
Answer (d) is equal to 27
Solution
Maths sir · 3 steps
Each sixth power is one big term plus one tiny term. Keep the big ones, and the x³ cancels cleanly.
- Spot the tiny terms. Rationalise: √(3x+1)-√(3x-1) = 2/(√(3x+1)+√(3x-1)) and x-√(x²-1) = 1/(x+√(x²-1)). Both go to 0 as x → ∞, so their sixth powers vanish.
- Size the big terms. For large x, √(3x+1)+√(3x-1) ≈ 2√3x, so its sixth power is about 64 × 27x³ = 1728x³. Also x+√(x²-1) ≈ 2x, so its sixth power is about 64x⁶.
lim(x → ∞) 1728x³/64x⁶ × x³ = 1728/64 = 27
- So. The limit is equal to 27, which is option (d).
The idea to keep. When a conjugate pair a+b and a-b appears and a-b → 0, only the a+b term matters. Divide everything by the highest power of x to see what survives.
Answer from the NTA final answer key. Solution by PrepWiser.