Question

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

Answer 1

Let $\alpha ,\beta ϵR$ be such that
${lim}_{x\to 0}\frac{x^{2}sin\left(\beta x\right)}{\alpha x-sinx}=1,\text{then}6(\alpha +\beta )$ equals
Using the infinite series expansion of sin x, this can be written as
${lim}_{x\to 0}\frac{x^2\{\beta x-\frac{(\beta x{)}^3}{3!}+\frac{(\beta x{)}^5}{5!}-\dots \}}{\alpha x-\{x-\frac{x^3}{3!}+\frac{x^5}{5!}-\dots \}}$
$\Rightarrow {lim}_{x\to 0}\frac{x^3\{\beta -\frac{{\beta }^3{x}^2}{3!}+\frac{{\beta }^5{x}^4}{5!}-\dots \}}{x(\alpha -1)+\{\frac{x^3}{3!}-\frac{x^5}{5!}+\dots \}}=1$
Since the value of the limit is equal to 1 we do not want the term $(\alpha -1)x$ to exist in the denominator. Hence $\alpha -1=0\to \alpha =1$
${lim}_{x\to 0}\frac{\text{/}{x}^3\{\beta -\frac{{\beta }^3{x}^2}{3!}+\frac{{\beta }^5{x}^4}{5!}\dots \}}{\text{/}{x}^3\{\frac{1}{3!}-\frac{x^2}{5!}+\dots \}}=1$
${lim}_{x\to 0}\left\{\frac{\beta -\frac{{\beta }^3{x}^2}{3!}+\frac{{\beta }^5{x}^4}{5!}-\dots }{\frac{1}{3!}-\frac{x^2}{5!}+\dots }\right\}=1$
$\Rightarrow \frac{\beta }{\frac{1}{6}}=1\Rightarrow \beta =\frac{1}{6}$
$6(\alpha +\beta )=6(1+\frac{1}{6})=7$