$$\mathop {\lim }\limits_{x \to \infty } \left( {\sqrt {{x^2} - x + 1} } \right) - ax = b$$ ($$\infty$$ $$-$$ $$\infty$$)

Now, $$\mathop {\lim }\limits_{x \to \infty } {{({x^2} - x + 1 - {a^2}{x^2}}) \over {\sqrt {{x^2} - x + 1} + ax}} = b$$

$$ \Rightarrow \mathop {\lim }\limits_{x \to \infty } {{(1 - {a^2}){x^2} - x + 1} \over {\sqrt {{x^2} - x + 1} + ax}} = b$$

$$ \Rightarrow \mathop {\lim }\limits_{x \to \infty } {{(1 - {a^2}){x^2} - x + 1} \over {x\left( {\sqrt {1 - {1 \over x} + {1 \over {{x^2}}}} + a} \right)}} = b$$

$$ \Rightarrow 1 - {a^2} = 0 \Rightarrow a = 1$$

Now, $$\mathop {\lim }\limits_{x \to \infty } {{ - x + 1} \over {x\left( {\sqrt {1 - {1 \over x} + {1 \over {{x^2}}}} + a} \right)}} = b$$

$$ \Rightarrow {{ - 1} \over {1 + a}} = b \Rightarrow b = - {1 \over 2}$$

$$(a,b) = \left( {1, - {1 \over 2}} \right)$$

$$\mathop {\lim }\limits_{x \to \beta } {{{e^{2({x^2} + bx + c)}} - 1 - 2({x^2} + bx + c)} \over {{{(x - \beta )}^2}}}$$

$$ = \mathop {\lim }\limits_{x \to \beta } {{1\left( {1 + {{2({x^2} + bx + c)} \over {1!}} + {{{2^2}{{({x^2} + bx + c)}^2}} \over {2!}} + ...} \right) - 1 - 2({x^2} + bx + c)} \over {{{(x - \beta )}^2}}}$$

$$ = \mathop {\lim }\limits_{x \to \beta } {{2{{({x^2} + bx + 1)}^2}} \over {{{(x - \beta )}^2}}}$$

$$ = \mathop {\lim }\limits_{x \to \beta } {{2{{(x - \alpha )}^2}{{(x - \beta )}^2}} \over {{{(x - \beta )}^2}}}$$

$$ = 2{(\beta - \alpha )^2} = 2({b^2} - 4c)$$

$$S = \mathop {\lim }\limits_{x \to 2} \sum\limits_{n = 1}^9 {{x \over {n(n + 1){x^2} + 2(2n + 1)x + 4}}} $$

$$S = \sum\limits_{n = 1}^9 {{2 \over {4({n^2} + 3n + 2)}}} = {1 \over 2}\sum\limits_{n = 1}^9 {\left( {{1 \over {n + 1}} - {1 \over {n + 2}}} \right)} $$

$$S = {1 \over 2}\left( {{1 \over 2} - {1 \over {11}}} \right) = {9 \over {44}}$$

$$L = \mathop {\lim }\limits_{n \to \infty } {1 \over n}\sum\limits_{r = 0}^{2n - 1} {{1 \over {1 + 4{{\left( {{r \over n}} \right)}^2}}}} $$

$$ \Rightarrow L = \int\limits_0^2 {{1 \over {1 + 4{x^2}}}dx} $$

$$ \Rightarrow L = \left. {{1 \over 2}{{\tan }^{ - 1}}(2x)} \right|_0^2 \Rightarrow L = {1 \over 2}{\tan ^{ - 1}}4$$