If $$\mathop {\lim }\limits_{x \to 0} {{(1 + {a^3}) + 8{e^{1/x}}} \over {1 + (1 - {b^3}){e^{1/x}}}} = 2$$, then

If the derivative of the function $$f(x) = \left\{ {\matrix{ {b{x^2} + ax + 4;} & {x \ge - 1} \cr {a{x^2} + b;} & {x < - 1} \cr } } \right.$$ is everywhere continuous, then

If $$\mathop {\lim }\limits_{x \to \infty } {\left( {1 + {a \over x} + {b \over {{x^2}}}} \right)^{2x}} = {e^2}$$, then

If $$L = \mathop {\lim }\limits_{x \to 0} {{a - \sqrt {{a^2} - {x^2}} - {{{x^2}} \over 4}} \over {{x^4}}},a > 0$$. If L is finite, then