$\lim _{x \rightarrow \infty} \left(\frac{x+a}{x+b}\right)^{x}$ is equal to

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

If $I = \lim_{x \rightarrow 0} \sin \left( \frac{e^{x}-x-1-\frac{x^{2}}{2}}{x^{2}} \right)$, then the limit

If $\alpha, \beta$ are the roots of the quadratic equation $ax^2 + bx + c = 0$,then $\lim_{x \to \alpha} \frac{1 - \cos(ax^2 + bx + c)}{(x - \alpha)^2}$ equals

$\lim _{x \rightarrow 1} \left( \lim _{y \rightarrow \infty} y \left( (e^x)^{1/y} - 1 \right) \right) = $

If $[x]$ denotes the greatest integer $\leq x$,then $\lim_{n \rightarrow \infty} \frac{1}{n^3} \{[1^2 x] + [2^2 x] + [3^2 x] + \ldots + [n^2 x] \} = $