Let beta = lim_{x to 0} frac{alpha x - (e^{3x | vedclass

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(C) Given $\beta = \lim_{x 	o 0} \frac{\alpha x - (e^{3x} - 1)}{\alpha x(e^{3x} - 1)}$.Using the expansion $e^{3x} = 1 + 3x + \frac{(3x)^2}{2!} + \do

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$\frac{5}{2}$

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(C) Given $\beta = \lim_{x 	o 0} \frac{\alpha x - (e^{3x} - 1)}{\alpha x(e^{3x} - 1)}$.Using the expansion $e^{3x} = 1 + 3x + \frac{(3x)^2}{2!} + \dots$,we have:$\beta = \lim_{x 	o 0} \frac{\alpha x - (1 + 3x + \frac{9x^2}{2} + \dots - 1)}{\alpha x(3x + \frac{9x^2}{2} + \dots)}$$\beta = \lim_{x 	o 0} \frac{(\alpha - 3)x - \frac{9}{2}x^2 - \dots}{3\alpha x^2 + \dots}$.For the limit to exist,the coefficient of $x$ in the numerator must be zero,so $\alpha - 3 = 0$,which gives $\alpha = 3$.Substituting $\alpha = 3$ into the expression:$\beta = \lim_{x 	o 0} \frac{-\frac{9}{2}x^2}{3(3)x^2} = \frac{-9/2}{9} = -\frac{1}{2}$.Finally,$\alpha + \beta = 3 - \frac{1}{2} = \frac{5}{2}$.

Let $\beta = \lim_{x \to 0} \frac{\alpha x - (e^{3x} - 1)}{\alpha x(e^{3x} - 1)}$ for some $\alpha \in R$. Then the value of $\alpha + \beta$ is:

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