$\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:

Let $l = \mathop {\lim}\limits_{x \to 0} \frac{[x]^2}{x^2}$ and $m = \mathop {\lim}\limits_{x \to 0} \frac{[x^2]}{x^2}$,where $[ \cdot ]$ denotes the greatest integer function. Then:

If $\lim _{x \rightarrow 2} \frac{1+\sqrt{1+4 \log _2 x}}{2+\left(2 x+\sin ^2 x+2 \cos x\right)(2 x-4)}=m$,then $m(m-1)=$

$\lim _{x \rightarrow \infty}\left(\frac{3 x^2-2 x+3}{3 x^2+x-2}\right)^{3 x-2} = $

For each $t \in \mathbb{R}$,let $[t]$ be the greatest integer less than or equal to $t$. Then $\lim_{x \to 1^+} \frac{(1 - |x| + \sin |1 - x|) \sin (\frac{\pi}{2} [1 - x])}{|1 - x|^2}$ is: