For alpha, beta, gamma in mathbbR$\alpha, \beta, \gamma \in \mathbb{R}$, if lim_x to 0 fracx^2 sin alpha x + (gamma - 1) e^x^2sin 2x - beta x = 3$\lim_{x \to 0} \frac{x^2 \sin \alpha x + (\gamma - 1) e^{x^2}}{\sin 2x - \beta x} = 3$, then beta + gamma - alpha$\beta + \gamma - \alpha$ is equal to:

### Related Formula Standard Taylor series expansions centered at x=0$x=0$: sin x = x - fracx^36 + dots$$\sin x = x - \frac{x^3}{6} + \dots$$ e^x = 1 + x + fracx^22 + dots$$e^x = 1 + x + \frac{x^2}{2} + \dots$$ ### Core Logic Since the limit evaluates to a finite value (3$3$) while the denominator goes to zero when x to 0$x \to 0$ (if 2-beta=0$2-\beta=0$), the numerator coefficients of lower-degree terms must vanish to resolve the indetermination. ### Step 1: Substitute Expansions Substitute series expansions into numerator and denominator: textNumerator = x^2(alpha x) + (gamma - 1)left(1 + x^2 + fracx^42 + dotsright)$$\text{Numerator} = x^2(\alpha x) + (\gamma - 1)\left(1 + x^2 + \frac{x^4}{2} + \dots\right)$$ textDenominator = left(2x - frac8x^36 + dotsright) - beta x = (2 - beta)x - frac43x^3 + dots$$\text{Denominator} = \left(2x - \frac{8x^3}{6} + \dots\right) - \beta x = (2 - \beta)x - \frac{4}{3}x^3 + \dots$$ ### Step 2: Equate Coefficients to Avoid Infinity Combine terms by degree: lim_x to 0 frac(gamma - 1) + (gamma - 1)x^2 + alpha x^3(2 - beta)x - frac43x^3 = 3$$\lim_{x \to 0} \frac{(\gamma - 1) + (\gamma - 1)x^2 + \alpha x^3}{(2 - \beta)x - \frac{4}{3}x^3} = 3$$ For a valid finite limit, the lowest power in the numerator cannot be smaller than the lowest power in the denominator. * Constraining constant term to zero: gamma - 1 = 0 implies gamma = 1$\gamma - 1 = 0 \implies \gamma = 1$ * This also forces the x^2$x^2$ coefficient to vanish: (gamma - 1) = 0$(\gamma - 1) = 0$. * To balance the remaining leading x^3$x^3$ terms, the x$x$ term in the denominator must vanish: 2 - beta = 0 implies beta = 2$2 - \beta = 0 \implies \beta = 2$. ### Step 3: Evaluate Remaining Limit Value Now compute the remaining simplified limit of x^3$x^3$ variables: lim_x to 0 fracalpha x^3-frac43x^3 = frac-3alpha4 = 3 implies alpha = -4$$\lim_{x \to 0} \frac{\alpha x^3}{-\frac{4}{3}x^3} = \frac{-3\alpha}{4} = 3 \implies \alpha = -4$$ ### Step 4: Final Expression Calculation Substitute the found parameters into beta + gamma - alpha$\beta + \gamma - \alpha$: beta + gamma - alpha = 2 + 1 - (-4) = 7$$\beta + \gamma - \alpha = 2 + 1 - (-4) = 7$$ ### Pattern Recognition Taylor expansions are far safer than consecutive L'Hopital iterations here because multiple variables are spread across distinct polynomial powers, isolating components explicitly by structural degree. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Limits, Continuity and Differentiability