The value of $\lim _{x \rightarrow 0} \left( \frac{1}{x} \ln \sqrt{\frac{1+x}{1-x}} \right)$ is

$\lim _{x \rightarrow \infty} x\left(\log \left(1+\frac{x}{2}\right)-\log \frac{x}{2}\right) = $

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

$\mathop {\lim }\limits_{x \to 0} \frac{{{{\left( {1 - \cos 2x} \right)}^2}}}{{2x\tan x - x\tan 2x}}$ is

$\lim _{x \rightarrow 0} \frac{1-\cos (1-\cos x)}{\sin ^4 x} = $

Let $f(x)=5-|x-2|$ and $g(x)=|x+1|, x \in R$. If $f(x)$ attains its maximum value at $\alpha$ and $g(x)$ attains its minimum value at $\beta$,then $\lim _{x \rightarrow-\alpha \beta} \frac{(x-1)\left(x^2-5 x+6\right)}{x^2-6 x+8}$ is equal to