$\mathop {\lim }\limits_{x \to \frac{\pi }{2}} (1 - \sin x)\tan x$ is

If $\lim _{x \rightarrow 0} \frac{ae^{x}-b \cos x + ce^{-x}}{x \sin x} = 2,$ then $a + b + c$ is equal to ...........

The value of the limit $\mathop {\lim }\limits_{x \to 0} {\left\{ {{1^{\frac{1}{{{{\sin }^2}x}}}} + {2^{\frac{1}{{{{\sin }^2}x}}}} + \dots + {n^{\frac{1}{{{{\sin }^2}x}}}}} \right\}^{{{\sin }^2}x}}$ is

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 \pi /2} (\sec \theta - \tan \theta ) = $

If $\alpha = \lim_{x \rightarrow 0} \frac{x \cdot 2^x - x}{1 - \cos x}$ and $\beta = \lim_{x \rightarrow 0} \frac{x \cdot 2^x - x}{\sqrt{1 + x^2} - \sqrt{1 - x^2}}$,then