Mathematics - Limits Question with Solution | TestHub

If $\alpha \ne 1,$ then $\underset{x\to 0}{\lim }\frac{x\sin \left(\beta x\right)}{\left(\frac{\alpha x-\sin x}{x}\right)}=0$
$\therefore$ $\alpha =1$
$\Rightarrow$ $\underset{x\to 0}{\lim }\frac{\beta x^3\left(\frac{\sin \left(\beta x\right)}{\beta x}\right)}{x^3\left(\frac{x-\sin x}{x^3}\right)}=\frac{\frac{\beta }{1}}{6}$

Using,$\underset{\mathrm{x}\to 0}{\lim }\frac{\mathrm{x}-\mathrm{sinx}}{{\mathrm{x}}^3}=\frac{1}{6} \&\underset{x\to 0}{\lim }\frac{\mathit{sinx}}{x}=1$
$\Rightarrow 6\beta =1\Rightarrow \beta =\frac{1}{6}$
$\Rightarrow 6\left(\alpha +\beta \right)=7$