$\mathop {\lim }\limits_{n \to \infty } {\left( {\frac{{{n^2} - n + 1}}{{{n^2} - n - 1}}} \right)^{n(n - 1)}} = $

Let $m$ and $n$ be two positive integers greater than $1$. If $\lim_{\alpha \rightarrow 0} \left( \frac{e^{\cos(\alpha^n)} - e}{\alpha^m} \right) = -\left( \frac{e}{2} \right)$,then the value of $\frac{m}{n}$ is

The quadratic equation whose roots are $m$ and $n$,where $m = \lim_{x \rightarrow 0} \frac{x \log(1+2x)}{x \tan x}$ and $n = \lim_{x \rightarrow 0} \frac{\log x + \log(\frac{1+x}{x})}{x}$,is

$\lim _{x}$ ${\rightarrow 0} \frac{8}{x^8}\left[1-\cos \left(\frac{x^2}{2}\right)-\cos \left(\frac{x^2}{4}\right)+\cos \left(\frac{x^2}{2}\right) \cdot \cos \left(\frac{x^2}{4}\right)\right]$ is equal to

$\lim _{x \rightarrow 0} \frac{\cos 4 x-4 \cos 2 x+3}{x^4} = $

$\mathop {Lim}\limits_{x \to \infty } \frac{2 + 2x + \sin 2x}{(2x + \sin 2x)e^{\sin x}}$ is :