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(B) Given the limit $\lim _{x \rightarrow 0} \frac{\frac{\alpha}{2} \int_0^x \frac{1}{1-t^2} d t+\beta x \cos x}{x^3} = 2$. Using the $L$'$H$ôpital's

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$2$

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(B) Given the limit $\lim _{x \rightarrow 0} \frac{\frac{\alpha}{2} \int_0^x \frac{1}{1-t^2} d t+\beta x \cos x}{x^3} = 2$. Using the $L$'$H$ôpital's rule,we differentiate the numerator and denominator with respect to $x$: $\lim _{x \rightarrow 0} \frac{\frac{\alpha}{2} \left(\frac{1}{1-x^2}\right) + \beta \cos x - \beta x \sin x}{3 x^2} = 2$. Using the Taylor series expansion for $\frac{1}{1-x^2} = 1+x^2+x^4+\dots$,$\cos x = 1-\frac{x^2}{2}+\dots$,and $\sin x = x-\frac{x^3}{6}+\dots$: $\lim _{x}$ ${\rightarrow 0} \frac{\frac{\alpha}{2}(1+x^2+x^4+\dots) + \beta(1-\frac{x^2}{2}+\dots) - \beta x(x-\frac{x^3}{6}+\dots)}{3 x^2} = 2$. Grouping terms by powers of $x$: $\lim _{x}$ ${\rightarrow 0} \frac{(\frac{\alpha}{2} + \beta) + x^2(\frac{\alpha}{2} - \frac{\beta}{2} - \beta) + O(x^4)}{3 x^2} = 2$. For the limit to exist and equal $2$,the constant term must be zero: $\frac{\alpha}{2} + \beta = 0 \implies \alpha = -2\beta$. Then the limit becomes $\lim _{x \rightarrow 0} \frac{x^2(\frac{\alpha}{2} - \frac{3\beta}{2})}{3 x^2} = \frac{\alpha - 3\beta}{6} = 2$. Substituting $\alpha = -2\beta$: $\frac{-2\beta - 3\beta}{6} = 2 \implies -5\beta = 12 \implies \beta = -2.4$. Then $\alpha = -2(-2.4) = 4.8$. Thus,$\alpha + \beta = 4.8 - 2.4 = 2.4$.

Let $\alpha$ and $\beta$ be real numbers such that $\lim _{x \rightarrow 0} \frac{1}{x^3}\left(\frac{\alpha}{2} \int_0^x \frac{1}{1-t^2} d t+\beta x \cos x\right)=2$. Then the value of $\alpha+\beta$ is $....$ (in $.40$)

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