JEE Main 2025 — 2 April, Morning Shift — Question 39

Given limit is of the form 0/0, so numerator must vanish at $x=0: (γ-1)e^0 = 0 ⇒ γ=1.$ Then limit becomes $\lim_{x\to 0} \frac{x^2 \sin(\alpha x)}{\sin(2x) - \beta x}$. Using series expansions: $\sin(\alpha x) = \alpha x - \frac{\alpha^3 x^3}{6} + \cdots$, $\sin(2x) = 2x - \frac{8x^3}{6} + \cdots$. Substitute: $\lim_{x\to 0} \frac{x^2(\alpha x - \frac{\alpha^3 x^3}{6} + \cdots)}{(2x - \frac{8x^3}{6} + \cdots) - \beta x} = \lim_{x\to 0} \frac{\alpha x^3 - \frac{\alpha^3 x^5}{6} + \cdots}{x(2-\beta) - \frac{8x^3}{6} + \cdots}$. For limit to exist and be finite, coefficient of x in denominator must be zero: 2-β=0 ⇒ β=2. Then limit becomes $\lim_{x\to 0} \frac{\alpha x^3 - \frac{\alpha^3 x^5}{6} + \cdots}{-\frac{8x^3}{6} + \cdots} = \frac{\alpha}{-8/6} = -\frac{3\alpha}{4}$. Set equal to 3: $-\frac{3\alpha}{4} = 3$ $⇒ α = -4.$ Thus $β+γ-α = 2+1-(-4) = 7.$