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Let α,β∈R\alpha , \beta \in \mathbb Rα,β∈R be such that limx→0x2sin(βx)αx−sinx=1\displaystyle{\lim_{x \rightarrow 0} \dfrac{x^2 \sin(\beta x)}{\alpha x - \sin x} = 1}x→0limαx−sinxx2sin(βx)=1. Then 6(α+β)+2=?6(\alpha + \beta)+2= ?6(α+β)+2=?
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