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(B) Given $\lim _{x \rightarrow 0} \frac{x^2 \sin (\beta x)}{\alpha x-\sin x}=1$.Using the Taylor series expansion for $\sin(x) = x - \frac{x^3}{3!} +

Vedclass · Accepted Answer

$7$

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(B) Given $\lim _{x \rightarrow 0} \frac{x^2 \sin (\beta x)}{\alpha x-\sin x}=1$.Using the Taylor series expansion for $\sin(x) = x - \frac{x^3}{3!} + \dots$ and $\sin(\beta x) = \beta x - \frac{(\beta x)^3}{3!} + \dots$,we get:$\lim _{x \rightarrow 0} \frac{x^2(\beta x - \frac{\beta^3 x^3}{6} + \dots)}{\alpha x - (x - \frac{x^3}{6} + \dots)} = 1$$\lim _{x \rightarrow 0} \frac{\beta x^3 - \frac{\beta^3 x^5}{6} + \dots}{(\alpha - 1)x + \frac{x^3}{6} - \dots} = 1$For the limit to be a finite non-zero value,the coefficient of $x$ in the denominator must be zero,so $\alpha - 1 = 0 \Rightarrow \alpha = 1$.Substituting $\alpha = 1$ into the limit:$\lim _{x \rightarrow 0} \frac{\beta x^3 - \frac{\beta^3 x^5}{6} + \dots}{\frac{x^3}{6} - \dots} = 1$Dividing numerator and denominator by $x^3$:$\lim _{x \rightarrow 0} \frac{\beta - \frac{\beta^3 x^2}{6} + \dots}{\frac{1}{6} - \dots} = 1$$\frac{\beta}{1/6} = 1$ $\Rightarrow 6\beta = 1$ $\Rightarrow \beta = \frac{1}{6}$.Therefore,$6(\alpha + \beta) = 6(1 + \frac{1}{6}) = 6 + 1 = 7$.

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

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