Find the limit: $\mathop {\lim }\limits_{x \to 3} [x(x+1)]$

$[x]$ denotes the greatest integer less than or equal to $x$. If $\{x\}=x-[x]$ and $\lim _{x \rightarrow 0^{-}} \frac{\sin ^{-1}(x+[x])}{2-\{x\}}=\theta$,then $\sin \theta+\cos \theta=$

$\lim _{x \rightarrow 0} \frac{\sqrt{2}-\sqrt{1+\cos x}}{\sqrt{15+\cos 2x}-4} = $

If $f(x) = \frac{2}{x - 3}$,$g(x) = \frac{x - 3}{x + 4}$ and $h(x) = - \frac{2(2x + 1)}{x^2 + x - 12}$,then $\lim_{x \to 3} [f(x) + g(x) + h(x)]$ is

Find the limit: $\mathop {\lim }\limits_{x \to 2} \left[\frac{x^{3}-2 x^{2}}{x^{2}-5 x+6}\right]$

$\lim_{x \rightarrow \frac{\pi}{2}} (\tan^{2} x (\sqrt{2 \sin^{2} x + 3 \sin x + 4} - \sqrt{\sin^{2} x + 6 \sin x + 2}))$ is equal to