$\lim _{x \rightarrow 0} \frac{x \tan 2x - 2x \tan x}{(1 - \cos 2x)^2}$ is

The limit $\lim _{x \rightarrow 1} \frac{\sqrt{1-\cos 2(x-1)}}{x-1}$

The value of $\mathop {\lim }\limits_{x \to \infty } (\sqrt {{a^2}{x^2} + ax + 1} - \sqrt {{a^2}{x^2} + 1})$ is

$ABC$ is an isosceles triangle inscribed in a circle of radius $r$. If $AB = AC$ and $h$ is the altitude from $A$ to $BC$,and $P$ is the perimeter of $ABC$,then $\mathop {\lim }\limits_{h \to 0} \frac{\Delta }{{{P^3}}}$ equals (where $\Delta$ is the area of the triangle).

$\mathop {\lim }\limits_{x \to 0} {\left( {\frac{{1 + 5{x^2}}}{{1 + 3{x^2}}}} \right)^{1/{x^2}}} = $

Let $f(x)=5-|x-2|$ and $g(x)=|x+1|$,$x \in R$. If $f(x)$ attains its maximum value at $\alpha$ and $g(x)$ attains its minimum value at $\beta$,then $\lim _{x \rightarrow-\alpha \beta} \frac{(x-1)(x^2-5x+6)}{(x^2-6x+8)}$ is equal to