$\lim _{x \rightarrow 1} \frac{\log x}{1-x} = $

$\lim _{x \rightarrow 0} \frac{\left(1+\frac{x}{2}\right)^{5 / 7}-1}{x} = $

$\mathop {\lim }\limits_{x \to 0} \frac{{\log _e}(1 + x)}{{3^x - 1}} = $

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:

$\operatorname{Lim}_{x \rightarrow 0} \frac{e-(1+2 x)^{\frac{1}{2 x}}}{x}$ is equal to :

Let $f(2) = 4$ and $f'(2) = 4$,then $\mathop {\lim }\limits_{x \to 2} \,\frac{{xf(2) - 2f(x)}}{{x - 2}}$ equals