If $$\alpha=\lim _\limits{x \rightarrow 0^{+}}\left(\frac{\mathrm{e}^{\sqrt{\tan x}}-\mathrm{e}^{\sqrt{x}}}{\sqrt{\tan x}-\sqrt{x}}\right)$$ and $\beta=\lim _\limits{x \rightarrow 0}(1+\sin x)^{\frac{1}{2} \cot x}$ are the roots of the quadratic equation $\mathrm{a} x^2+\mathrm{b} x-\sqrt{\mathrm{e}}=0$, then $12 \log _{\mathrm{e}}(\mathrm{a}+\mathrm{b})$ is equal to _________.

<p>$$\begin{aligned} \alpha & =\lim _{x \rightarrow 0^{+}} \frac{e^{\sqrt{\tan x}}-e^{\sqrt{x}}}{(\sqrt{\tan x}-\sqrt{x})} \\ & =\lim _{x \rightarrow 0} \frac{e^{\sqrt{x}}\left(e^{\sqrt{\tan x}-\sqrt{x}}-1\right)}{(\sqrt{\tan x}-\sqrt{x})}=1 \\ \beta & =\lim _{x \rightarrow 0}(1+\sin x)^{\frac{1}{2} \cot x}=\lim _{x \rightarrow 0} e^{(\sin x)\left(\frac{1}{2} \cot x\right)} \\ & =\lim _{x \rightarrow 0} e^{\frac{1}{2} \cos x}=e^{1 / 2} \end{aligned}$$</p> <p>$$\begin{aligned} & \text { Product of roots }=\sqrt{e}=\frac{-\sqrt{e}}{a} \Rightarrow a=-1 \\ & \text { Sum of roots }=\frac{-b}{a}=1+\sqrt{e} \\ & \qquad=b=\sqrt{e}+1 \\ & \Rightarrow 12 \ln (a+b)=12 \ln (\sqrt{e}+1-1)=12 \ln \left(e^{1 / 2}\right)=6 \end{aligned}$$</p>