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

The value of $\mathop {\lim }\limits_{x \to 0} \frac{1}{x}\left[ {{{\tan }^{ - 1}}\left( {\frac{{x + 1}}{{2x + 1}}} \right) - \frac{\pi }{4}} \right]$ is

If $l, m$ $(l < m)$ are roots of $ax^2 + bx + c = 0$,then $\lim_{x \rightarrow \alpha} \frac{|ax^2 + bx + c|}{ax^2 + bx + c} = $

$\lim _{x \rightarrow 0} \frac{\sqrt{1+x^2}-\sqrt{1-x+x^2}}{3^x-1}$ is equal to

$\lim _{x \rightarrow \frac{\pi}{2}} \frac{(1-\sin x)(8 x^3-\pi^3) \cos x}{(\pi-2 x)^4}$

$\lim _{x \rightarrow \infty} x\left(\log \left(1+\frac{x}{2}\right)-\log \frac{x}{2}\right) = $