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

If $\lim _{x \rightarrow 0} \frac{e^{2 x}-1}{5 x}=l$ and $\lim _{x \rightarrow 1} \frac{2}{x-1} \log x=m$,then the cubic equation whose roots are $5l, m$,and $1$ is:

The value of $\mathop {\lim }\limits_{n \to \infty } {\left( {e \cdot {a^2} \cdot {e^3} \cdot {a^4} \cdots {e^{n - 1}} \cdot {a^n}} \right)^{\frac{1}{{{n^2} + 1}}}}$ is equal to

The value of $\mathop {\text{Limit}}\limits_{x \to \infty } \frac{{\cot ^{ - 1}}\left( {{x^{ - a}}\log _a x} \right)}{{\sec ^{ - 1}}\left( {{a^x}\log _x a} \right)}$ for $a > 1$ is equal to

$\lim _{x \rightarrow 0} \frac{\sqrt{1+x^2}-\sqrt{1-x+x^2}}{3^x-1}$ 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 $m$.