JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

Let L = _ x 0 x^2 ^2 x x^2 - ^2 x Dividing the numerator and the denominator by x^4, we get: L = _ x 0 ( x x )^2 x^2 - ^2 x x^4 Factorizing the denominator: L = _ x 0 ( x x )^2 ( x - x x^3 ) ( x + x x ) Using the standard limits: _ x 0 x x = 1 _ x 0 x - x x^3 = 1 6 (using Taylor series expansion x = x - x^3 3! + ) _ x 0 x + x x = _ x 0 (1 + x x ) = 1 + 1 = 2 Substituting these limits into the expression: L = 1^2 ( 1 6 ) 2 = 1 1 3 = 3 Answer: 3