The value of the limit of $\frac{x^3 - 8}{x^2 - 4}$ as $x \to 2$ is

$\mathop {\lim }\limits_{x \to 0} \frac{{x({2^x} - 1)}}{{1 - \cos x}} = $

$\lim _{x \rightarrow 0} \frac{e^{x^2}-\cos 3 x}{\sin x \log (1+2 x)}=$

The quadratic equation whose roots are $\ell = \lim_{\theta \rightarrow 0} \left( \frac{3 \sin \theta - 4 \sin^3 \theta}{\theta} \right)$ and $m = \lim_{\theta \rightarrow 0} \left( \frac{2 \tan \theta}{\theta(1 - \tan^2 \theta)} \right)$ is

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

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