Evaluate: $\mathop {\lim }\limits_{x \to 0} \frac{\sqrt{1+x}-1}{x}$

$\mathop {\lim }\limits_{x \to 0} \frac{{{e^{1/x}} - 1}}{{{e^{1/x}} + 1}} = $

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

$\lim_{x \rightarrow -2^{+}} ([x]^2 - [x] - 2) + \lim_{x \rightarrow -3^{-}} ([x]^2 - 4[x] + 3) =$

$\lim _{\theta \rightarrow \frac{\pi}{2}^{-}} \frac{8 \tan ^4 \theta+4 \tan ^2 \theta+5}{(3-2 \tan \theta)^4} = $

For $\mathop {\text{Lim}}\limits_{x \to 8} \,\,\frac{{\sin \{ x - 10\} }}{{\{ 10 - x\} }}$ (where $\{ \}$ denotes the fractional part function),determine the existence of the limits.