$\mathop {\lim }\limits_{x \to 0} \left[ {\frac{1}{x} - \frac{{\log (1 + x)}}{{{x^2}}}} \right] =$

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:

$\mathop {\lim }\limits_{\theta \to {0^ + }} {(\sin \theta )^{(\sin \theta - {{\sin }^2}\theta )}}$ is

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

Let $f: R \rightarrow R$ be a function such that $f(2)=4$ and $f^{\prime}(2)=1$. Then,the value of $\lim _{x \rightarrow 2} \frac{x^{2} f(2)-4 f(x)}{x-2}$ is equal to:

If $f$ is a real function such that $f(4)=4$ and $f^{\prime}(4)=16$,then $\lim _{x \rightarrow 4} \frac{\sqrt{f(x)}-2}{\sqrt{x}-2} =$