Cho hàm số \({\rm{f}}\left( {\rm{x}} \right) = \left\{ {\begin{array}{*{20}{c}}{{{\rm{x}}^2} - 3\,\,{\rm{khi}}\,\,{\rm{x}} \ge 2}\\{{\rm{x}} - 1\,\,{\rm{khi}}\,\,{\rm{x}} < 2}\end{array}} \right.\). Chọn kết quả đúng của \[\mathop {\lim }\limits_{{\rm{x}} \to 2} {\rm{f}}\left( {\rm{x}} \right)\]

Ta có \[\mathop {\lim }\limits_{{\rm{x}} \to 1} \frac{{\sqrt {{{\rm{x}}^2} + {\rm{x}} + 2} - \sqrt[3]{{7{\rm{x}} + 1}}}}{{\sqrt 2 \left( {{\rm{x}} - 1} \right)}} = \mathop {\lim }\limits_{{\rm{x}} \to 1} \frac{{\sqrt {{{\rm{x}}^2} + {\rm{x}} + 2} - 2 + 2 - \sqrt[3]{{7{\rm{x}} + 1}}}}{{\sqrt 2 \left( {{\rm{x}} - 1} \right)}}\]

\[ = \mathop {\lim }\limits_{{\rm{x}} \to 1} \frac{{\sqrt {{{\rm{x}}^2} + {\rm{x}} + 2} - 2}}{{\sqrt 2 \left( {{\rm{x}} - 1} \right)}} + \mathop {\lim }\limits_{{\rm{x}} \to 1} \frac{{2 - \sqrt[3]{{7{\rm{x}} + 1}}}}{{\sqrt 2 \left( {{\rm{x}} - 1} \right)}} = {\rm{I + J}}\]

Tính\[{\rm{I}} = \mathop {\lim }\limits_{{\rm{x}} \to 1} \frac{{\sqrt {{{\rm{x}}^2} + {\rm{x}} + 2} - 2}}{{\sqrt 2 \left( {{\rm{x}} - 1} \right)}} = \mathop {\lim }\limits_{{\rm{x}} \to 1} \frac{{{{\rm{x}}^2} + {\rm{x}} + 2 - 4}}{{\sqrt 2 \left( {{\rm{x}} - 1} \right)\left( {\sqrt {{{\rm{x}}^2} + {\rm{x}} + 2} + 2} \right)}}\]

\[ = \mathop {\lim }\limits_{{\rm{x}} \to 1} \frac{{\left( {{\rm{x}} - 1} \right)\left( {{\rm{x}} + 2} \right)}}{{\sqrt 2 \left( {{\rm{x}} - 1} \right)\left( {\sqrt {{{\rm{x}}^2} + {\rm{x}} + 2} + 2} \right)}} = \mathop {\lim }\limits_{{\rm{x}} \to 1} \frac{{{\rm{x}} + 2}}{{\sqrt 2 \left( {\sqrt {{{\rm{x}}^2} + {\rm{x}} + 2} + 2} \right)}} = \frac{3}{{4\sqrt 2 }}\]

và\[{\rm{J}} = \mathop {\lim }\limits_{{\rm{x}} \to 1} \frac{{2 - \sqrt[3]{{7{\rm{x}} + 1}}}}{{\sqrt 2 \left( {{\rm{x}} - 1} \right)}} = \mathop {\lim }\limits_{{\rm{x}} \to 1} \frac{{8 - 7{\rm{x}} - 1}}{{\sqrt 2 \left( {{\rm{x}} - 1} \right)\left[ {4 + 2\sqrt[3]{{7{\rm{x}} + 1}} + {{\left( {\sqrt[3]{{7{\rm{x}} + 1}}} \right)}^2}} \right]}}\]

\[ = \mathop {\lim }\limits_{{\rm{x}} \to 1} \frac{{ - 7}}{{\sqrt 2 \left[ {4 + 2\sqrt[3]{{7{\rm{x}} + 1}} + {{\left( {\sqrt[3]{{7{\rm{x}} + 1}}} \right)}^2}} \right]}} = \frac{{ - 7}}{{12\sqrt 2 }}\]

Do đó\[\mathop {\lim }\limits_{{\rm{x}} \to 1} \frac{{\sqrt {{{\rm{x}}^2} + {\rm{x}} + 2} - \sqrt[3]{{7{\rm{x}} + 1}}}}{{\sqrt 2 \left( {{\rm{x}} - 1} \right)}} = {\rm{I + J}} = \frac{{\sqrt 2 }}{{12}}\]

Suy ra a = 1, b = 12, c = 0. Vậy a + b + c = 13.

Đáp án cần chọn là: A

\[\mathop {\lim }\limits_{{\rm{x}} \to - \infty } \left( {\sqrt {9{{\rm{x}}^2} - {\rm{ax}}} + \sqrt[3]{{27{{\rm{x}}^3} + {\rm{b}}{{\rm{x}}^2} + 5}}} \right) = \mathop {\lim }\limits_{{\rm{x}} \to - \infty } \left( {\sqrt {9{{\rm{x}}^2} - {\rm{ax}}} + {\rm{3x}} + \sqrt[3]{{27{{\rm{x}}^3} + {\rm{b}}{{\rm{x}}^2} + 5}} - 3{\rm{x}}} \right)\]

\[ = \mathop {\lim }\limits_{{\rm{x}} \to - \infty } \left( {\sqrt {9{{\rm{x}}^2} - {\rm{ax}}} + 3{\rm{x}}} \right) + \mathop {\lim }\limits_{{\rm{x}} \to - \infty } \left( {\sqrt[3]{{{\rm{27}}{{\rm{x}}^{\rm{3}}}{\rm{ + b}}{{\rm{x}}^{\rm{2}}}{\rm{ + 5}}}} - 3{\rm{x}}} \right)\]

Ta có :

\[\mathop {\lim }\limits_{{\rm{x}} \to - \infty } \left( {\sqrt {9{{\rm{x}}^2} - {\rm{ax}}} + 3{\rm{x}}} \right) = \mathop {\lim }\limits_{{\rm{x}} \to - \infty } \frac{{ - {\rm{ax}}}}{{\sqrt {9{{\rm{x}}^2} - {\rm{ax}}} - 3{\rm{x}}}} = \mathop {\lim }\limits_{{\rm{x}} \to - \infty } \frac{{\rm{a}}}{{\sqrt {9 - \frac{{\rm{a}}}{{\rm{x}}}} + 3}} = \frac{{\rm{a}}}{{\rm{6}}}\]

Ta có :

\[\mathop {\lim }\limits_{{\rm{x}} \to - \infty } \left( {\sqrt[3]{{{\rm{27}}{{\rm{x}}^{\rm{3}}}{\rm{ + b}}{{\rm{x}}^{\rm{2}}}{\rm{ + 5}}}} - {\rm{3x}}} \right)\]

\[ = \mathop {\lim }\limits_{{\rm{x}} \to - \infty } \frac{{\left( {\sqrt[3]{{{\rm{27}}{{\rm{x}}^{\rm{3}}}{\rm{ + b}}{{\rm{x}}^{\rm{2}}}{\rm{ + 5}}}} - 3{\rm{x}}} \right)\left( {\sqrt[3]{{{{\left( {{\rm{27}}{{\rm{x}}^{\rm{3}}}{\rm{ + b}}{{\rm{x}}^{\rm{2}}}{\rm{ + 5}}} \right)}^2}}} + 3{\rm{x}}.\sqrt[3]{{{\rm{27}}{{\rm{x}}^{\rm{3}}}{\rm{ + b}}{{\rm{x}}^{\rm{2}}}{\rm{ + 5}}}} + 9{{\rm{x}}^2}} \right)}}{{\sqrt[3]{{{{\left( {{\rm{27}}{{\rm{x}}^{\rm{3}}}{\rm{ + b}}{{\rm{x}}^{\rm{2}}}{\rm{ + 5}}} \right)}^2}}} + 3{\rm{x}}.\sqrt[3]{{{\rm{27}}{{\rm{x}}^{\rm{3}}}{\rm{ + b}}{{\rm{x}}^{\rm{2}}}{\rm{ + 5}}}} + 9{{\rm{x}}^2}}}\]

\[ = \mathop {\lim }\limits_{{\rm{x}} \to - \infty } \frac{{{\rm{b}}{{\rm{x}}^{\rm{2}}}{\rm{ + 5}}}}{{\sqrt[3]{{{{\left( {{\rm{27}}{{\rm{x}}^{\rm{3}}}{\rm{ + b}}{{\rm{x}}^{\rm{2}}}{\rm{ + 5}}} \right)}^2}}} + 3{\rm{x}}.\sqrt[3]{{{\rm{27}}{{\rm{x}}^{\rm{3}}}{\rm{ + b}}{{\rm{x}}^{\rm{2}}}{\rm{ + 5}}}} + 9{{\rm{x}}^2}}}\]

\[ = \mathop {\lim }\limits_{{\rm{x}} \to - \infty } \frac{{{\rm{b + }}\frac{{\rm{5}}}{{{{\rm{x}}^{\rm{2}}}}}}}{{\sqrt[3]{{{{\left( {{\rm{27 + }}\frac{{\rm{b}}}{{\rm{x}}}{\rm{ + }}\frac{{\rm{5}}}{{{{\rm{x}}^{\rm{3}}}}}} \right)}^2}}} + 3.\sqrt[3]{{{\rm{27 + }}\frac{{\rm{b}}}{{\rm{x}}}{\rm{ + }}\frac{{\rm{5}}}{{{{\rm{x}}^{\rm{2}}}}}}} + 9}} = \frac{{\rm{b}}}{{{\rm{27}}}}\]

Do đó\[\frac{{\rm{a}}}{{\rm{6}}}{\rm{ + }}\frac{{\rm{b}}}{{{\rm{27}}}}{\rm{ = }}\frac{{\rm{7}}}{{{\rm{27}}}}\]

Áp dụng bất đẳng thức Cauchy cho 2 số dương, ta có :\[\frac{{\rm{a}}}{{\rm{6}}}{\rm{ + }}\frac{{\rm{b}}}{{{\rm{27}}}} \ge 2\sqrt {\frac{{\rm{a}}}{{\rm{6}}}{\rm{.}}\frac{{\rm{b}}}{{{\rm{27}}}}} \]

\[ \Rightarrow \frac{7}{{27}} \ge \frac{2}{{9\sqrt 2 }}\sqrt {{\rm{a}}{\rm{.b}}} \Rightarrow {\rm{ab}} \le \frac{{49}}{{18}}\]

Đẳng thức xảy ra khi\(\left\{ {\begin{array}{*{20}{c}}{\frac{{\rm{a}}}{{\rm{6}}}{\rm{ = }}\frac{{\rm{b}}}{{{\rm{2}}7}}}\\{\frac{{\rm{a}}}{{\rm{6}}} + \frac{{\rm{b}}}{{{\rm{2}}7}} = \frac{{\rm{7}}}{{{\rm{2}}7}}}\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{{\rm{a}} = \frac{7}{9}}\\{{\rm{b}} = \frac{7}{2}}\end{array}} \right.\)

Vậy giá trị lớn nhất của ab bằng \[\frac{{49}}{{18}}\].

Đáp án cần chọn là: B