$\mathop {\lim }\limits_{n \to \infty } \frac{{\sqrt n }}{{\sqrt n + \sqrt {n + 1} }} = $

The quadratic polynomial $p(x)$ has roots $1$ and $\alpha$, while quadratic polynomial $q(x)$ has roots $1$ and $\beta$. Let $\alpha$ and $\beta$ be the roots of $r(x) = p(x) + q(x)$. Then $\lim_{x \to \infty} [\sqrt{p(x)} - \sqrt{q(x)}] = $

$\lim \limits _{x \to 0} \frac{{{{(\sin x - \tan x)}^2} - {{(1 - \cos 2x)}^4} + {x^5}}}{{7\cdot{{({{\tan }^{ - 1}}x)}^7}\, + {{({{\sin }^{ - 1}}x)}^6}+ 3{{\sin }^5}x}}$ is equal to

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

$\mathop {\lim }\limits_{n \to \infty } {({3^n} + {4^n})^{\frac{1}{n}}} = $

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