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

Let f(x) = 5 – |x – 2| and g(x) = |x + 1|, x $$ \in $$ R. If f(x) attains maximum value at $$\alpha $$ and g(x) attains minimum value at $$\beta $$, then $$\mathop {\lim }\limits_{x \to -\alpha \beta } {{\left( {x - 1} \right)\left( {{x^2} - 5x + 6} \right)} \over {{x^2} - 6x + 8}}$$ is equal to :

If $$\alpha $$ and $$\beta $$ are the roots of the equation 375x² – 25x – 2 = 0, then $$\mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 1}^n {{\alpha ^r}} + \mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 1}^n {{\beta ^r}} $$ is equal to :

If $$\mathop {\lim }\limits_{x \to 1} {{{x^2} - ax + b} \over {x - 1}} = 5$$, then a + b is equal to :