Question:

Rearrange and group the terms in the numerator of $g(x)$: $$3xe^x + 3x - \alpha e^x - \alpha x = 3x(e^x + 1) - \alpha(e^x + x)$$ Now divide by the denominator $3(e^x+1)$ to simplify $g(x)$: $$g(x) = \dfrac{3x(e^x + 1) - \alpha(e^x + x)}{3(e^x + 1)} = x - \dfrac{\alpha}{3} \cdot \dfrac{e^x + x}{e^x + 1}$$ We need to find the limit: $$L = \lim_{x \to x_0} \dfrac{g(x) + e^{x_0}}{x - x_0}$$ Since $e^{x_0} + x_0 = 0 \implies e^{x_0} = -x_0$. Substitute this into the limit: $$L = \lim_{x \to x_0} \dfrac{x - \dfrac{\alpha}{3} \cdot \dfrac{e^x + x}{e^x + 1} - x_0}{x - x_0}$$ $$L = \lim_{x \to x_0} \left( 1 - \dfrac{\alpha}{3(e^x + 1)} \cdot \dfrac{e^x + x}{x - x_0} \right)$$ Since $e^{x_0} + x_0 = 0$, we can write $\dfrac{e^x + x}{x - x_0}$ as $\dfrac{(e^x + x) - (e^{x_0} + x_0)}{x - x_0}$. Taking the limit as $x \to x_0$ gives the derivative of $(e^x + x)$ at $x_0$: $$\lim_{x \to x_0} \dfrac{e^x + x - (e^{x_0} + x_0)}{x - x_0} = \left. \dfrac{d}{dx}(e^x + x) \right|_{x=x_0} = e^{x_0} + 1$$ Substituting this back into the limit expression: $$L = 1 - \dfrac{\alpha}{3(e^{x_0} + 1)} \cdot (e^{x_0} + 1) = 1 - \dfrac{\alpha}{3}$$ Thus, the absolute value of the limit is $\left|1 - \dfrac{\alpha}{3}\right|$: - For $\alpha = 2$: the limit is $\left|1 - \dfrac{2}{3}\right| = \dfrac{1}{3}$. - For $\alpha = 3$: the limit is $\left|1 - \dfrac{3}{3}\right| = 0$. Hence, statement C is true.