Biết  \(\mathop {\lim }\limits_{x \to 8} \frac{{\...

C

- Hướng dẫn giải

Phương pháp giải:

Đặt \(t = x - 8 \Rightarrow x = t + 8 \Rightarrow \mathop {\lim }\limits_{x \to 8} \;t = \mathop {\lim }\limits_{x \to 8} \;\left( {x - 8} \right) = 0.\)   Khi đó ta có:

\(\frac{{\sqrt {x + 1}  - \sqrt[3]{{x + 19}}}}{{\sqrt[4]{{x + 8}} - 2}} = \frac{{\sqrt {t + 9}  - \sqrt[3]{{t + 27}}}}{{\sqrt[4]{{t + 16}} - 2}} = \frac{{3\sqrt {1 + \frac{t}{9}}  - 3\sqrt[3]{{1 + \frac{t}{{27}}}}}}{{2\sqrt[4]{{1 + \frac{t}{{16}}}} - 2}} = \frac{3}{2}\,.\,\frac{{\frac{{\sqrt {1 + \frac{t}{9}}  - 1}}{t} - \frac{{\sqrt[3]{{1 + \frac{t}{{27}}}} - 1}}{t}}}{{\frac{{\sqrt[4]{{1 + \frac{t}{{16}}}} - 1}}{t}}} = f\left( t \right)\)

Áp dụng công thức:  \(\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt[n]{{1 + ax}} - 1}}{x} = \frac{a}{n}\;\left( {a \ne 0;\;\;n \in {\mathbb{N}^*}} \right).\)

Giải chi tiết:

Đặt \(t = x - 8 \Rightarrow x = t + 8 \Rightarrow \mathop {\lim }\limits_{x \to 8} \;t = \mathop {\lim }\limits_{x \to 8} \;\left( {x - 8} \right) = 0.\)   Khi đó ta có:

\(f\left( t \right) = \frac{{\sqrt {t + 9}  - \sqrt[3]{{t + 27}}}}{{\sqrt[4]{{t + 16}} - 2}} = \frac{{3\sqrt {1 + \frac{t}{9}}  - 3\sqrt[3]{{1 + \frac{t}{{27}}}}}}{{2\sqrt[4]{{1 + \frac{t}{{16}}}} - 2}} = \frac{3}{2}\,.\,\frac{{\frac{{\sqrt {1 + \frac{t}{9}}  - 1}}{t} - \frac{{\sqrt[3]{{1 + \frac{t}{{27}}}} - 1}}{t}}}{{\frac{{\sqrt[4]{{1 + \frac{t}{{16}}}} - 1}}{t}}}.\)

Do đó: \(\mathop {\lim }\limits_{x \to 8} \frac{{\sqrt {x + 1}  - \sqrt[3]{{x + 19}}}}{{\sqrt[4]{{x + 8}} - 2}} = \mathop {\lim }\limits_{t \to 0} f\left( t \right) = \frac{3}{2}\,\mathop {\lim }\limits_{t \to 0} \frac{{\frac{{\sqrt {1 + \frac{t}{9}}  - 1}}{t} - \frac{{\sqrt[3]{{1 + \frac{t}{{27}}}} - 1}}{t}}}{{\frac{{\sqrt[4]{{1 + \frac{t}{{16}}}} - 1}}{t}}}\) .

Ta có:

Xét:  \(\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt[n]{{1 + ax}} - 1}}{x} = \frac{a}{n}\;\left( {a \ne 0;\;\;n \in {\mathbb{N}^*}} \right).\)

Đặt \(\sqrt[n]{{1 + ax}} = y \Leftrightarrow {y^n} = 1 + ax \Rightarrow x = \frac{{{y^n} - 1}}{a};\,\;\,\mathop {\lim }\limits_{x \to 0} y = 1\)                                                                                                   

\(\begin{array}{l}\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt[n]{{1 + ax}} - 1}}{x} = a\mathop {\lim }\limits_{y \to 1} \frac{{y - 1}}{{{y^n} - 1}} = a\mathop {\lim }\limits_{y \to 1} \frac{{y - 1}}{{\left( {y - 1} \right)\left( {{y^{n - 1}} + ... + y + 1} \right)}} = a\mathop {\lim }\limits_{y \to 1} \frac{1}{{{y^{n - 1}} + ... + y + 1}} = \frac{a}{n}\\\mathop {\lim }\limits_{t \to 0} \frac{{\sqrt {1 + \frac{t}{9}}  - 1}}{t} = \frac{{\frac{1}{9}}}{2} = \frac{1}{{18}};\;\;\;\mathop {\lim }\limits_{t \to 0} \frac{{\sqrt[3]{{1 + \frac{t}{{27}}}} - 1}}{t} = \,\frac{{\frac{1}{{27}}}}{3} = \frac{1}{{81}};\;\;\;\mathop {\lim }\limits_{t \to 0} \frac{{\sqrt[4]{{1 + \frac{t}{{16}}}} - 1}}{t} = \frac{{\frac{1}{{16}}}}{4} = \frac{1}{{64}}.\\ \Rightarrow \mathop {\lim }\limits_{t \to 0} f\left( t \right) = \frac{3}{2}.\frac{{\frac{1}{{18}} - \frac{1}{{81}}}}{{\frac{1}{{64}}}} = \frac{{112}}{{27}}.\\ \Rightarrow \mathop {\lim }\limits_{x \to 8} \frac{{\sqrt {x + 1}  - \sqrt[3]{{x + 19}}}}{{\sqrt[4]{{x + 8}} - 2}} = \frac{{112}}{{27}}.\\ \Rightarrow \left\{ \begin{array}{l}a = 112\\b = 27\end{array} \right. \Rightarrow a + b = 139.\end{array}\)

Chọn C.