<s>[INST] if $\lim_{x\to 0} \frac{x^2\sin(\beta x)}{\alpha x-\sin x} = 1$ then $6(\alpha+\beta)=?$ 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 what is the value of $6(\alpha+\beta)$
 [/INST] $$
\begin{aligned}
\lim _{x\to 0}\left(\frac{x^2\sin \left(\beta \:x\right)}{\alpha \:x-\sin \:x}\right)
& = \lim _{x\to 0}\left(\frac{x^2\left(\beta x+o\left(x\right)\right)}{\alpha \:x-\left(x-\frac{x^3}{3!}+o(x^3)\right)}\right)
\\& = \beta \:\cdot \lim _{x\to 0}\left(\frac{6x^2}{6α+x^2-6}\right) (\text{so } \color{red}{\beta \ne 0})
\\& = \beta \:\cdot\lim _{x\to \:0}\left(\frac{6x^2}{6+x^2-6}\right) = 6\beta   \text{ (with } \color{red}{\alpha = 1})
\\& = \color{red}{\frac{1}{6}} \:\cdot\lim _{x\to \:0}\left(\frac{6x^2}{6+x^2-6}\right) = \color{blue}{1}   \text{ (with } \color{red}{\beta = \frac{1}{6}})
\end{aligned}
$$
So, with
$$\alpha = 1, \beta = \frac{1}{6}$$
$$6(\alpha + \beta)=6\left(1+\frac{1}{6}\right) = \color{gold}{7}$$
</s>