$$ \mathrm{e}^{\mathrm{x}_0}+\mathrm{x}_0=0 $$

$$ \begin{aligned} & g(x)=\frac{3 x e^x+3 x-\alpha e^x-\alpha x}{3\left(e^x+1\right)} \\ & g(x)=x-\frac{\alpha\left(e^x+x\right)}{3\left(e^x+1\right)}, g(x)+e^{x_0}=x+e^{x_0}-\frac{\alpha}{3}\left(\frac{e^x+x}{e^x+1}\right) \end{aligned} $$

$$ g(x)+e^{x_0}=\left(x-x_0\right)-\frac{\alpha}{3}\left(\frac{e^x+x}{e^x+1}\right) $$

$\Rightarrow$ जैसा कि स्पष्ट है, $\mathrm{g}\left(\mathrm{x}_0\right)+\mathrm{x}_0 \Rightarrow 0$

$$ \lim \limits_{x \rightarrow x_0}\left|\frac{g(x)+e^{x_0}}{x-x_0}\right| \frac{0 \uparrow}{0 \uparrow}=\lim \limits_{x \rightarrow x_0}\left|\frac{g^{\prime}(x)}{1}\right|=\left|g^{\prime}\left(x_0\right)\right| $$

$$ \mathrm{g}^{\prime}(\mathrm{x})=1-\frac{\alpha}{3}\left(\frac{\left(\mathrm{e}^{\mathrm{x}}+1\right)\left(\mathrm{e}^{\mathrm{x}}+1\right)-\left(\mathrm{e}^{\mathrm{x}}+\mathrm{x}\right) \mathrm{e}^{\mathrm{x}}}{\left(\mathrm{e}^{\mathrm{x}}+1\right)^2}\right) $$

$$ \mathrm{g}^{\prime}\left(\mathrm{x}_0\right)=1-\frac{\alpha}{3}\left(\frac{\left(\mathrm{e}^{\mathrm{x}_0}+1\right)^2}{\left(\mathrm{e}^{\mathrm{x}_0}+1\right)^2}\right)=1-\frac{\alpha}{3} $$

$$ \Rightarrow \lim \limits_{x \rightarrow x_0}\left|\frac{g(x)+e^{x_0}}{x-x_0}\right|=\left|g^{\prime}\left(x_0\right)\right|=\left|1-\frac{\alpha}{3}\right| $$