Let $\alpha, \beta \in \mathbb{R}$ be such that $\lim_{x \to 0}\frac{x^{2}\sin(\beta x)}{\alpha x-\sin x}=1$ then $6(\alpha+\beta)$ equals.

$\alpha,\beta\in\mathbb{R}, \quad \lim_{x\to 0}\dfrac{x^{2}\sin(\beta x)}{\alpha x-\sin x}=1$ $\text{To find: } 6(\alpha+\beta)$ $\text{numerator is } O(x^3) \Rightarrow \text{denominator must be } O(x^3)$ $\text{expand } \sin x \text{ and } \sin(\beta x)$ $\text{match coefficients} \Rightarrow \alpha,\beta$ $\sin x=x-\dfrac{x^3}{6}+\dfrac{x^5}{120}-\cdots$ $\alpha x-\sin x=(\alpha-1)x+\dfrac{x^3}{6}-\dfrac{x^5}{120}+\cdots$ $\alpha-1\ne 0 \Rightarrow \text{denominator} \sim x \Rightarrow \text{limit } 0 \text{ or } \infty$ $\alpha-1=0 \Rightarrow \alpha=1$ $\alpha x-\sin x=\dfrac{x^3}{6}-\dfrac{x^5}{120}+\cdots\sim \dfrac{x^3}{6}$ $\sin(\beta x)=\beta x-\dfrac{(\beta x)^3}{6}+\cdots$ $x^2\sin(\beta x)=\beta x^3-\dfrac{\beta^3 x^5}{6}+\cdots\sim \beta x^3$ $\lim_{x\to 0}\dfrac{x^2\sin(\beta x)}{\alpha x-\sin x}=\dfrac{\beta x^3}{x^3/6}=6\beta$ $6\beta=1 \Rightarrow \beta=\dfrac{1}{6}$ $\alpha=1, \quad \beta=\dfrac{1}{6}$ $\dfrac{x^2\sin(x/6)}{x-\sin x}\to \dfrac{x^3/6}{x^3/6}=1\ \checkmark$ $6(\alpha+\beta)=6\left(1+\dfrac{1}{6}\right)=6\cdot\dfrac{7}{6}=7$ $\boxed{6(\alpha+\beta)=7}$