$\mathop {\lim }\limits_{x \to \infty } \frac{{(x - 1)(2x + 3)}}{{{x^2}}} = $

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

$\lim_{x \rightarrow -\infty} \frac{3|x|-x}{|x|-2x} - \lim_{x \rightarrow 0} \frac{\log(1+x^3)}{\sin^3 x} =$

The value of $\mathop {\lim }\limits_{x \to 2} \frac{{{3^{x/2}} - 3}}{{{3^x} - 9}}$ is

$\lim _{x \rightarrow 0} \frac{\sin |x|}{x}$ is equal to

If $l = \lim_{\theta \rightarrow 0} \left( \frac{3 \sin \theta - 4 \sin^3 \theta}{\theta} \right)$ and $m = \lim_{\theta \rightarrow 0} \left( \frac{2 \tan \theta}{\theta(1 - \tan^2 \theta)} \right)$,find the quadratic equation whose roots are $l$ and $m$.