For α, β, γ ∈ R, if lim x → 0 (x² sin α x+(γ-1) e^(x²))/(sin 2 x-β x)=3, then β+γ-α is equal to :

Solution

Maths sir · 4 steps

Expand both sides in powers of x, then make the low powers vanish one at a time until the limit can equal 3.

1. Kill the constant. As x → 0 the denominator goes to 0, but the numerator goes to γ - 1. For the limit to be a finite 3, the numerator must also go to 0, so γ = 1. The numerator is then just x² sin α x ≈ α x³.
2. Match the powers. Using sin 2x = 2x - 8x³/6 + …, the denominator is (2-β)x - 4/3x³ + …. If β ≠ 2, it behaves like x while the numerator behaves like x³, so the limit would be 0. Hence β = 2.

   lim(x → 0) (α x³ + …)/(-4/3x³ + …) = -3α/4 = 3

3. Find alpha. -3α/4 = 3 gives α = -4.
4. So. β + γ - α = 2 + 1 - (-4) = 7, which is option (d).

The idea to keep. When a limit with unknown constants must be finite and nonzero, expand top and bottom as series. Make the lower powers cancel until both start at the same power of x.

Answer from the NTA final answer key. Solution by PrepWiser.