$$\mathop {\lim }\limits_{x \to {\pi \over 2}} {{\cot x - \cos x} \over {{{\left( {\pi - 2x} \right)}^3}}}$$ equals

$$\mathop {\lim }\limits_{x \to 0} \,{{{{\left( {1 - \cos 2x} \right)}^2}} \over {2x\,\tan x\, - x\tan 2x}}$$ is :

Let a, b $$ \in $$ R, (a $$ \ne $$ 0). If the function f defined as

$$f\left( x \right) = \left\{ {\matrix{ {{{2{x^2}} \over a}\,\,,} & {0 \le x < 1} \cr {a\,\,\,,} & {1 \le x < \sqrt 2 } \cr {{{2{b^2} - 4b} \over {{x^3}}},} & {\sqrt 2 \le x < \infty } \cr } } \right.$$

is continuous in the interval [0, $$\infty $$), then an ordered pair ( a, b) is :

If    $$\mathop {\lim }\limits_{x \to \infty } {\left( {1 + {a \over x} - {4 \over {{x^2}}}} \right)^{2x}} = {e^3},$$ then 'a' is equal to :