Mathematics - Limits Question with Solution | TestHub

$\begin{array}{rl}& e^{x_0}+x_0=0\\ & g(x)=\frac{3x{e}^x+3x-\alpha e^x-\alpha x}{3\left(e^x+1\right)}\\ & g(x)=x-\frac{\alpha \left(e^x+x\right)}{3\left(e^x+1\right)},g(x)+e^{x_0}=x+e^{x_0}-\frac{\alpha }{3}\left(\frac{e^x+x}{e^x+1}\right)\\ & g(x)+e^{x_0}=\left(x-x_0\right)-\frac{\alpha }{3}\left(\frac{e^x+x}{e^x+1}\right)\\ & \Rightarrow \text{ As very clear }g\left(x_0\right)+x_0\Rightarrow 0\end{array}$ $\begin{array}{rl}& \underset{x\to x_0}{\lim }\left|\frac{g(x)+e^{x_0}}{x-x_0}\right|\frac{0\uparrow }{0\uparrow }=\underset{x\to x_0}{\lim }\left|\frac{g^{\prime }(x)}{1}\right|=\left|g^{\prime }\left(x_0\right)\right|\\ & g^{\prime }(x)=1-\frac{\alpha }{3}\left(\frac{\left(e^x+1\right)\left(e^x+1\right)-\left(e^x+x\right)e^x}{{\left(e^x+1\right)}^2}\right)\\ & g^{\prime }\left(x_0\right)=1-\frac{\alpha }{3}\left(\frac{{\left(e^{x_0}+1\right)}^2}{{\left(e^{x_0}+1\right)}^2}\right)=1-\frac{\alpha }{3}\\ & \Rightarrow \underset{x\to x_0}{\lim }\left|\frac{g(x)+e^{x_0}}{x-x_0}\right|=\left|g^{\prime }\left(x_0\right)\right|=\left|1-\frac{\alpha }{3}\right|\end{array}$