Question

Let $\alpha ,\beta \in R$ be such that $\underset{x\to 0}{lim}\frac{x^2\sin \left(\beta x\right)}{\alpha x-\sin x}=1.$ Then $6(\alpha +\beta )$ equals

Answer 1

$\underset{x\to 0}{lim}\frac{x^2\sin \left(\beta x\right)}{\alpha x-\sin x}=1\underset{x\to 0}{lim}\frac{x^2(\beta x-\frac{{\beta }^3{x}^3}{3!}+....)}{\alpha x-(x-\frac{x^3}{3!}+....)}=1\underset{x\to 0}{lim}\frac{x^3(\beta -\frac{{\beta }^3{x}^2}{3!}+....)}{(\alpha -1)x+\frac{x^3}{3!}-....}=1\underset{x\to 0}{lim}\frac{x^2(\beta -\frac{{\beta }^3{x}^2}{3!}+....)}{(\alpha -1)+\frac{x^2}{3!}-....}=1\Rightarrow \alpha -1=0\Rightarrow \alpha =1\text{At}\alpha =1\underset{x\to 0}{lim}\frac{x^2(\beta -\frac{{\beta }^3{x}^2}{3!}+....)}{\frac{x^2}{3!}-....}=1\Rightarrow \beta \times 3!=1\Rightarrow \beta =\frac{1}{6}\therefore 6(\alpha +\beta )=6(1+\frac{1}{6})=7$