Biết mathop lim limitsx to 8 căn x + 1 - căn [3]x + 19căn [4]x + 8
![Biết mathop lim limitsx to 8 căn x + 1 - căn [3]x + 19căn [4]x + 8](https://tuhoc365.vn/wp-content/uploads/2020/03/qa-238x145.png "Biết mathop lim limitsx to 8 căn x + 1 - căn [3]x + 19căn [4]x + 8")
Câu hỏi
Nhận biếtBiết \( \mathop { \lim } \limits_{x \to 8} \frac{{ \sqrt {x + 1} - \sqrt[3]{{x + 19}}}}{{ \sqrt[4]{{x + 8}} - 2}} = \frac{a}{b} \) trong đó \( \frac{a}{b} \) là phân số tối giản, \(a \) và \(b \) là các số nguyên dương. Tổng \(a + b \) bằng
Đáp án đúng: C
Lời giải của Tự Học 365
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.
