$\mathop {\lim }\limits_{x \to \pi /6} \left[ {\frac{{3\sin x - \sqrt 3 \cos x}}{{6x - \pi }}} \right] = $

$\mathop {\lim }\limits_{x \to 0} \frac{{{2^x} - 1}}{{{{(1 + x)}^{1/2}} - 1}} = $

Let $f : R \rightarrow R$ be a differentiable function such that $f \left(\frac{\pi}{4}\right)=\sqrt{2}$,$f \left(\frac{\pi}{2}\right)=0$ and $f^{\prime}\left(\frac{\pi}{2}\right)=1$. Let $g(x)=\int\limits_{x}^{\pi / 4}\left(f^{\prime}(t) \sec t+\tan t \sec t f(t)\right) d t$ for $x \in\left[\frac{\pi}{4}, \frac{\pi}{2}\right)$. Then $\lim\limits _{ x \rightarrow\left(\frac{\pi}{2}\right)^{-}} g ( x )$ is equal to

$\lim _{x \rightarrow 0} \left( \frac{3^{x}-1}{x} \right)$ is equal to

$\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 + \sin x} - \sqrt {1 - \sin x} }}{x} = $

The value of $\lim _{x \rightarrow 0} \frac{1}{x^3} \int_0^x \frac{t \ln (1+t)}{t^4+4} dt$ is