Question 1

Given $e^{x_0} + x_0 = 0 \implies e^{x_0} = -x_0$. --- (1)

We can rewrite $g(x)$ as:

$g(x) = \frac{3x(e^x + 1) - \alpha(e^x + x)}{3(e^x + 1)} = x - \frac{\alpha}{3} \cdot \frac{e^x + x}{e^x + 1}$ ----(2)

For $x = x_0$, from equations (1) and (2) we have $g(x_0) = x_0$.

Also, $e^{x_0} = -x_0$, so $g(x_0) + e^{x_0} = x_0 - x_0 = 0$.

The limit $L = \lim_{x \to x_0} \left| \frac{g(x) + e^{x_0}}{x - x_0} \right| = \lim_{x \to x_0} \left| \frac{g(x) - g(x_0)}{x - x_0} \right| = |g'(x_0)|$.

Differentiating $g(x)$:

$g'(x) = 1 - \frac{\alpha}{3} \cdot \frac{(e^x + 1)^2 - (e^x + x)e^x}{(e^x + 1)^2}$

At $x = x_0$, $e^{x_0} + x_0 = 0$, so:

$g'(x_0) = 1 - \frac{\alpha}{3} \cdot \frac{(e^{x_0} + 1)^2 - 0}{(e^{x_0} + 1)^2} = 1 - \frac{\alpha}{3}$.

For $\alpha = 3$, $g'(x_0) = 1 - \frac{3}{3} = 0$.

Thus, for $\alpha = 3$, the limit is $0$.