$\mathop {\lim }\limits_{x \to \infty } (\sqrt {{x^2} + 1} - x)$ is equal to

$ABC$ 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).

$\lim _{x \rightarrow 1} \frac{(2 x-3)(\sqrt{x}-1)}{2 x^2+x-3}$

$\lim _{x \rightarrow 0} \frac{3^{\sin x}-2^{\tan x}}{\sin x}=$

$\mathop {\lim }\limits_{x \to 0} \sin \left( {\frac{1}{x}} \right)$ is

If $L = \lim_{x^2 \to a} \frac{b - \cos(x^2 - a)}{(x^2 - a) \sin(c(x^2 - a))}$ is a non-zero finite value $(a > 0)$,then: