a) Tính giới hạn \(\lim \left( {\sqrt {9{n^2} + 3n - 2} - 3n} \right).\)

       b) Cho hàm số \[f\left( x \right) = \left\{ {\begin{array}{*{20}{l}}{\frac{{x - \sqrt {x + 2} }}{{{x^2} - 4}}}&{{\rm{khi}}}&{x > 2}\\{{x^2} + ax + 3b}&{{\rm{khi}}}&{x < 2}\\{2a + b - 6}&{{\rm{khi}}}&{x = 2}\end{array}} \right.\] liên tục tại \[x = 2\].  Tính \[I = a + b\].

a)                                                      Ta có \(\lim \left( {\sqrt {9{n^2} + 3n - 2} - 3n} \right) = \lim \frac{{\left( {\sqrt {9{n^2} + 3n - 2} - 3n} \right)\left( {\sqrt {9{n^2} + 3n - 2} + 3n} \right)}}{{\sqrt {9{n^2} + 3n - 2} + 3n}}\)

\( = \lim \frac{{3n - 2}}{{\sqrt {9{n^2} + 3n - 2} + 3n}}\)\( = \lim \frac{{3 - \frac{2}{n}}}{{\sqrt {9 + \frac{3}{n} - \frac{2}{{{n^2}}}} + 3}} = \frac{1}{2}\).

Vậy \(\lim \left( {\sqrt {9{n^2} + 3n - 2} - 3n} \right) = \frac{1}{2}\).

b)                                                     Để hàm \[f\left( x \right)\] liên tục tại \[x = 2\] cần có \[\mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right) = f\left( 2 \right)\]

+) Ta có \[\mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ + }} \left( {\frac{{x - \sqrt {x + 2} }}{{{x^2} - 4}}} \right) = \mathop {\lim }\limits_{x \to {2^ + }} \frac{{{x^2} - x - 2}}{{\left( {x + 2} \right)\left( {x - 2} \right)\left( {x + \sqrt {x + 2} } \right)}}\]

\[ = \mathop {\lim }\limits_{x \to {2^ + }} \frac{{x + 1}}{{\left( {x + 2} \right)\left( {x + \sqrt {x + 2} } \right)}} = \frac{3}{{16}}\].

+) \[\mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ - }} \left( {{x^2} + ax + 3b} \right) = 2a + 3b + 4\]

+) \[f\left( 2 \right) = 2a + b - 6\]

Ta được hệ phương trình: \[\left\{ \begin{array}{l}2a + b - 6 = \frac{3}{{16}}\\2a + 3b + 4 = \frac{3}{{16}}\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a = \frac{{179}}{{32}}\\b = - 5\end{array} \right. \Rightarrow a + b = \frac{{19}}{{32}}\].