If lim _{x rightarrow 0} frac{3+alpha sin x+b | vedclass

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(C) Given $\lim _{x ightarrow 0} \frac{3+\alpha \sin x+\beta \cos x+\log _e(1-x)}{3 	an ^2 x}=\frac{1}{3}$.Using Taylor series expansions for $\sin

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$5$

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(C) Given $\lim _{x ightarrow 0} \frac{3+\alpha \sin x+\beta \cos x+\log _e(1-x)}{3 	an ^2 x}=\frac{1}{3}$.Using Taylor series expansions for $\sin x$,$\cos x$,and $\log _e(1-x)$ near $x=0$:$\sin x = x - \frac{x^3}{6} + \dots$$\cos x = 1 - \frac{x^2}{2} + \dots$$\log _e(1-x) = -x - \frac{x^2}{2} - \frac{x^3}{3} - \dots$Substituting these into the numerator:$3 + \alpha(x - \frac{x^3}{6}) + \beta(1 - \frac{x^2}{2}) + (-x - \frac{x^2}{2} - \frac{x^3}{3}) = (3+\beta) + (\alpha-1)x - (\frac{\beta+1}{2})x^2 + \dots$For the limit to exist and equal $\frac{1}{3}$,the coefficients of $x^0$ and $x^1$ must be zero:$3+\beta = 0 \Rightarrow \beta = -3$$\alpha-1 = 0 \Rightarrow \alpha = 1$Now,the limit becomes $\lim _{x ightarrow 0} \frac{-(\frac{-3+1}{2})x^2}{3x^2} = \frac{1}{3} 	imes \frac{2}{3} = \frac{1}{3}$,which is consistent.Finally,$2\alpha - \beta = 2(1) - (-3) = 2 + 3 = 5$.

If $\lim _{x \rightarrow 0} \frac{3+\alpha \sin x+\beta \cos x+\log _e(1-x)}{3 \tan ^2 x}=\frac{1}{3}$,then $2 \alpha-\beta$ is equal to :

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