$\mathop {\lim }\limits_{x \to {0^ + }} \left\{ {{{\left( {1 + x} \right)}^{\frac{2}{x}}}} \right\}$ is equal to (where $\{.\}$ denotes the fractional part of $x$)

The value of $\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{\left[ {1 - \tan \left( {\frac{x}{2}} \right)} \right]\,[1 - \sin x]}}{{\left[ {1 + \tan \left( {\frac{x}{2}} \right)} \right]\,{{[\pi - 2x]}^3}}}$ is

Let $a_1, a_2, a_3, \ldots, a_n$ be $n$ positive consecutive terms of an arithmetic progression. If $d > 0$ is its common difference,then evaluate $\lim_{n \rightarrow \infty} \sqrt{\frac{d}{n}} \left( \frac{1}{\sqrt{a_1} + \sqrt{a_2}} + \frac{1}{\sqrt{a_2} + \sqrt{a_3}} + \ldots + \frac{1}{\sqrt{a_{n-1}} + \sqrt{a_n}} \right)$.

$\lim _{x \rightarrow 1} \frac{2^{2 x-2}-2^x+1}{\sin ^2(x-1)}=$

$\lim _{x \rightarrow \infty}\left(\frac{x+6}{x+1}\right)^{x+4}$ is equal to

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: