$\lim\limits_{x \rightarrow 0} \frac{\int\limits_{0}^{x} t \sin (10 t) d t}{x}$ is equal to

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

$\mathop {\lim }\limits_{x \to 0} \frac{{\tan 2x - x}}{{3x - \sin x}} = $

If $l_1 = \lim_{x \rightarrow 2^{+}} (x + [x])$,$l_2 = \lim_{x \rightarrow 2^{-}} (2x - [x])$ and $l_3 = \lim_{x \rightarrow \pi/2} \frac{\cos x}{x - \pi/2}$,then:

If $f(1) = 1$ and $f'(1) = 4,$ then the value of $\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {f(x)} - 1}}{{\sqrt x - 1}}$ is

$\lim _{x \rightarrow 0} \left[ \frac{1}{x} - \frac{1}{e^x - 1} \right] = $