If $\lim _\limits{x \rightarrow \infty}\left(\left(\frac{\mathrm{e}}{1-\mathrm{e}}\right)\left(\frac{1}{\mathrm{e}}-\frac{x}{1+x}\right)\right)^x=\alpha$, then the value of $\frac{\log _{\mathrm{e}} \alpha}{1+\log _{\mathrm{e}} \alpha}$ equals :

If $\sum_\limits{r=1}^n T_r=\frac{(2 n-1)(2 n+1)(2 n+3)(2 n+5)}{64}$, then $\lim _\limits{n \rightarrow \infty} \sum_\limits{r=1}^n\left(\frac{1}{T_r}\right)$ is equal to :

$$\lim _\limits{x \rightarrow 0} \frac{e-(1+2 x)^{\frac{1}{2 x}}}{x}$$ is equal to

For $$\mathrm{a}, \mathrm{b}>0$$, let $$f(x)= \begin{cases}\frac{\tan ((\mathrm{a}+1) x)+\mathrm{b} \tan x}{x}, & x< 0 \\ 3, & x=0 \\ \frac{\sqrt{\mathrm{a} x+\mathrm{b}^2 x^2}-\sqrt{\mathrm{a} x}}{\mathrm{~b} \sqrt{\mathrm{a}} x \sqrt{x}}, & x> 0\end{cases}$$ be a continuous function at $$x=0$$. Then $$\frac{\mathrm{b}}{\mathrm{a}}$$ is equal to :