$\mathop {\lim }\limits_{x \to \infty } \frac{{(2x - 3)(3x - 4)}}{{(4x - 5)(5x - 6)}} = $

The quadratic equation whose roots are $l$ and $m$,where
$\begin{aligned}
& l=\lim _{\theta \rightarrow 0}\left(\frac{3 \sin \theta-4 \sin ^2 \theta}{\theta}\right), \\
& m=\lim _{\theta \rightarrow 0} \frac{2 \tan \theta}{\theta\left(1-\tan ^2 \theta\right)}, \text{ is}
\end{aligned}$

$\lim _{n \rightarrow \infty} \frac{1}{n^3} \sum_{k=1}^n k^2 x = $

$\mathop {Limit}\limits_{x \to \frac{\pi }{2}} \,\frac{{\sin x}}{{{{\cos }^{ - 1}}\left[ {\frac{1}{4}\,(3\sin x\, - \,\sin 3x)} \right]}}\,$,where $[ \cdot ]$ denotes the greatest integer function,is

$\mathop {Lim}\limits_{x \to {0^ - }} \sin^{-1}([\tan x])$ $= l$,then $\{l\}$ is equal to,where $[\cdot]$ and $\{\cdot\}$ denote the greatest integer and fractional part functions,respectively.

$\lim _{x \rightarrow \infty}\left[\frac{8 x^2+5 x+3}{2 x^2-7 x-5}\right]^{\frac{4 x+3}{8 x-1}} = $