$\mathop {\lim }\limits_{x \to \pi /2} \frac{\tan 3x}{x} = $

$\lim _{x \rightarrow 0} \frac{\sqrt{1-\cos x^2}}{1-\cos x} = $

$\lim_{x \rightarrow \frac{\pi}{2}} (\tan^{2} x (\sqrt{2 \sin^{2} x + 3 \sin x + 4} - \sqrt{\sin^{2} x + 6 \sin x + 2}))$ 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 _{y \rightarrow 0} \frac{\sqrt{1+\sqrt{1+y^4}}-\sqrt{2}}{y^4} = $

$\lim _{x \rightarrow 0} \frac{6^x-3^x-2^x+1}{x^2}$ is equal to