Let α, β ∈ R be such that limx→0 x²sin(βx) / (αx - sinx) = 1. Then 6(α + β) equals

limx→0 x²sin(βx) / (αx - sinx) = limx→0 x²(βx - β³x³/3! + β⁵x⁵/5! - …)/(αx - (x - x³/3! + x⁵/5! - …)) = limx→0 x³(β - β³x²/3! + β⁵x⁴/5! - …)/(x(α - 1) + x³/3! - x⁵/5! + …)= 1

Since we have no term of x in the numerator we need to put the coefficient of the term containing x in the denominator equal to 0, otherwise after limiting condition i.e. x = 0 is put we get indeterminate form. Now after the condition on α we can take the x³ common from numerator and denominator and then put x = 0 to get the desired value.

It gives, α - 1 = 0 ⇒ α = 1

Limit = β/1 = 1 ⇒ β = 1

6(α + β) = 6(1 + 1) = 12