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:

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

If $f(x)$ is a differentiable function of $x$,then $\mathop {Limit}\limits_{h \to 0} \frac{f(x + 3h) - f(x - 2h)}{h} = $

Which of the following limits vanish? (where $[ \cdot ]$ denotes the greatest integer function)

Let $g(x) = \frac{(x-1)^n}{\log \cos^m(x-1)}$ for $x \neq 1$,and let $p$ be the left-hand derivative of $|x-1|$ at $x=1$. If $\lim_{x \rightarrow 1^{+}} g(x) = p$,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