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

$\underset{x\to 0}{lim}\frac{x^2\sin \beta x}{\alpha x-\sin x}=\underset{x\to 0}{lim}\frac{x^2(\beta x-\frac{{\beta }^3{x}^3}{3!}+\frac{{\beta }^5{x}^5}{5!}-\cdots )}{\alpha x-(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots )}$

$=\underset{x\to 0}{lim}\frac{x^3(\beta -\frac{{\beta }^3{x}^2}{3!}+\frac{{\beta }^5{x}^4}{5!}-\cdots )}{(\alpha -1)x+\frac{x^3}{3!}-\frac{x^5}{5!}+\cdots }=1$

Since we have no term of $x$ in the numerator we need to put the coefficent of the term containing $x$ in the denominator equal to $0$ other wise after limiting condition i.e $x=0$ is put we get indeterminate form. Now after the condition on $\alpha$ we can take the $x^3$ common from numerator and denominator and then put $x=0$ to get the desired value.

It gives, $\alpha -1=0\Rightarrow \alpha =1$

Limit $=6\beta =1\Rightarrow \beta =\frac{1}{6}$

$\Rightarrow 6(\alpha +\beta )=6(1+\frac{1}{6})=7$