$$\mathop {\lim }\limits_{x \to 0} {{{{\sin }^2}x} \over {\sqrt 2 - \sqrt {1 + \cos x} }}$$ equals:

$$\mathop {\lim }\limits_{x \to {1^ - }} {{\sqrt \pi - \sqrt {2{{\sin }^{ - 1}}x} } \over {\sqrt {1 - x} }}$$ is equal to :

Let f be a differentiable function such that f(1) = 2 and f '(x) = f(x) for all x $$ \in $$ R R. If h(x) = f(f(x)), then h'(1) is equal to :

$$\mathop {\lim }\limits_{x \to \pi /4} {{{{\cot }^3}x - \tan x} \over {\cos \left( {x + {\pi \over 4}} \right)}}$$ is :