$\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {\frac{1}{2}(1 - \cos 2x)} }}{x} = $

If $\alpha$ is the interior angle of a regular octagon,then $\lim_{\theta \to \alpha^+} \frac{\tan \theta - 1}{[\sin \theta + \cos \theta]}$ is equal to (Note: $[k]$ denotes the greatest integer less than or equal to $k$).

The value of the limit of $\frac{x^3 - 8}{x^2 - 4}$ as $x \to 2$ is

If $0 < x < \frac{\pi}{2}$,then

Evaluate the limit: $\lim _{x}$ ${\rightarrow \infty} \frac{(\sqrt{3 x+1}+\sqrt{3 x-1})^6+(\sqrt{3 x+1}-\sqrt{3 x-1})^6}{\left(x+\sqrt{x^2-1}\right)^6+\left(x-\sqrt{x^2-1}\right)^6} x^3$

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