$\mathop {\lim }\limits_{x \to 1} f(x)$ is equal to,where $f(x) = \begin{cases} \frac{e^{\frac{1}{x-1}} - 2}{e^{\frac{1}{x-1}} + 2} & x \neq 1 \\ 1 & x = 1 \end{cases}$

Let $A = \lim_{x \rightarrow 0^{+}} \left(1 + \tan^2 \sqrt{x}\right)^{\frac{1}{2x}}$,then $\log_{e} A = $

If $a > 0$,$[\cdot]$ denotes the greatest integer function,$\lim _{x \rightarrow a^{-}}\left(\frac{|x|^3}{a}-\left[\frac{x}{a}\right]^3\right)=k$,and $\lim _{x \rightarrow a^{+}}\left(\frac{|x|^3}{a}-\left[\frac{x}{a}\right]^3\right)=l$,then:

The value of $\lim _{x \rightarrow 1} \frac{x+x^2+\ldots+x^n-n}{x-1}$ is

$\mathop {\lim }\limits_{n \to \infty } \,\frac{{\sum\limits_{r = 0}^n {{{\tan }^{ - 1}}\left( {1 + r + {r^2}} \right)} }}{n}$ is equal to

$\lim _{x \rightarrow \infty} x^3 \left[ \sqrt{x^2 + \sqrt{x^4 + 1}} - \sqrt{2} x \right] = $