Tính $\mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt {1 + 2x} .\sqrt[3]{{1 + 3x}}.\sqrt[4]{{1 + 4x}} - 1}}{x}$
-
A.
$\dfrac{{23}}{2}$.
-
B.
$24$.
-
C.
$\dfrac{3}{2}$.
-
D.
$3$.
- Biến đổi biểu thức, đưa về dạng $\mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt[n]{{1 + nx}} - 1}}{x}$
- Nhân liên hợp.
Ta có:
$\begin{array}{l}\sqrt {1 + 2x} .\sqrt[3]{{1 + 3x}}.\sqrt[4]{{1 + 4x}} - 1\\ = \sqrt {1 + 2x} - \sqrt {1 + 2x} + \sqrt {1 + 2x} .\sqrt[3]{{1 + 3x}} - \sqrt {1 + 2x} .\sqrt[3]{{1 + 3x}} + \sqrt {1 + 2x} .\sqrt[3]{{1 + 3x}}.\sqrt[4]{{1 + 4x}} - 1\\ = \left( {\sqrt {1 + 2x} - 1} \right) + \sqrt {1 + 2x} \left( {\sqrt[3]{{1 + 3x}} - 1} \right) + \sqrt {1 + 2x} .\sqrt[3]{{1 + 3x}}.\left( {\sqrt[4]{{1 + 4x}} - 1} \right)\end{array}$
$\begin{array}{l} \Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt {1 + 2x} .\sqrt[3]{{1 + 3x}}.\sqrt[4]{{1 + 4x}} - 1}}{x}\\ = \mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{\sqrt {1 + 2x} - 1}}{x}} \right) + \mathop {\lim }\limits_{x \to 0} \left( {\sqrt {1 + 2x} .\dfrac{{\sqrt[3]{{1 + 3x}} - 1}}{x}} \right) + \mathop {\lim }\limits_{x \to 0} \left( {\sqrt {1 + 2x} .\sqrt[3]{{1 + 3x}}.\dfrac{{\sqrt[4]{{1 + 4x}} - 1}}{x}} \right)\end{array}$
Tính:
$\mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{\sqrt {1 + 2x} - 1}}{x}} \right) = \mathop {\lim }\limits_{x \to 0} \dfrac{{\left( {\sqrt {1 + 2x} - 1} \right)\left( {\sqrt {1 + 2x} + 1} \right)}}{{x\left( {\sqrt {1 + 2x} + 1} \right)}} = \mathop {\lim }\limits_{x \to 0} \dfrac{{2x}}{{x\left( {\sqrt {1 + 2x} + 1} \right)}} = \mathop {\lim }\limits_{x \to 0} \dfrac{2}{{\sqrt {1 + 2x} + 1}} = \dfrac{2}{{1 + 1}} = 1$
$\begin{array}{l}\mathop {\lim }\limits_{x \to 0} \left( {\sqrt {1 + 2x} .\dfrac{{\sqrt[3]{{1 + 3x}} - 1}}{x}} \right) = \mathop {\lim }\limits_{x \to 0} \left( {\sqrt {1 + 2x} .\dfrac{{\left( {\sqrt[3]{{1 + 3x}} - 1} \right)\left[ {{{\left( {\sqrt[3]{{1 + 3x}}} \right)}^2} + \sqrt[3]{{1 + 3x}} + 1} \right]}}{{x.\left[ {{{\left( {\sqrt[3]{{1 + 3x}}} \right)}^2} + \sqrt[3]{{1 + 3x}} + 1} \right]}}} \right)\\ = \mathop {\lim }\limits_{x \to 0} \left( {\sqrt {1 + 2x} .\dfrac{{3x}}{{x.\left[ {{{\left( {\sqrt[3]{{1 + 3x}}} \right)}^2} + \sqrt[3]{{1 + 3x}} + 1} \right]}}} \right) = \mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{3\sqrt {1 + 2x} }}{{\left[ {{{\left( {\sqrt[3]{{1 + 3x}}} \right)}^2} + \sqrt[3]{{1 + 3x}} + 1} \right]}}} \right) = \dfrac{{3.1}}{{1 + 1 + 1}} = 1\end{array}$
$\begin{array}{l}\mathop {\lim }\limits_{x \to 0} \left( {\sqrt {1 + 2x} .\sqrt[3]{{1 + 3x}}.\dfrac{{\sqrt[4]{{1 + 4x}} - 1}}{x}} \right) = \mathop {\lim }\limits_{x \to 0} \left( {\sqrt {1 + 2x} .\sqrt[3]{{1 + 3x}}.\dfrac{{\left( {\sqrt[4]{{1 + 4x}} - 1} \right)\left[ {{{\left( {\sqrt[4]{{1 + 4x}}} \right)}^3} + {{\left( {\sqrt[4]{{1 + 4x}}} \right)}^2} + \sqrt[4]{{1 + 4x}} + 1} \right]}}{{x\left[ {{{\left( {\sqrt[4]{{1 + 4x}}} \right)}^3} + {{\left( {\sqrt[4]{{1 + 4x}}} \right)}^2} + \sqrt[4]{{1 + 4x}} + 1} \right]}}} \right)\\ = \mathop {\lim }\limits_{x \to 0} \left( {\sqrt {1 + 2x} .\sqrt[3]{{1 + 3x}}.\dfrac{{4x}}{{x\left[ {{{\left( {\sqrt[4]{{1 + 4x}}} \right)}^3} + {{\left( {\sqrt[4]{{1 + 4x}}} \right)}^2} + \sqrt[4]{{1 + 4x}} + 1} \right]}}} \right)\\ = \mathop {\lim }\limits_{x \to 0} \dfrac{{4\sqrt {1 + 2x} .\sqrt[3]{{1 + 3x}}}}{{{{\left( {\sqrt[4]{{1 + 4x}}} \right)}^3} + {{\left( {\sqrt[4]{{1 + 4x}}} \right)}^2} + \sqrt[4]{{1 + 4x}} + 1}} = \dfrac{{4.1.1}}{{1 + 1 + 1 + 1}} = 1\end{array}$
Vậy $\mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt {1 + 2x} .\sqrt[3]{{1 + 3x}}.\sqrt[4]{{1 + 4x}} - 1}}{x} = 1 + 1 + 1 = 3$
Đáp án : D
Các bài tập cùng chuyên đề
Tính $\mathop {\lim }\limits_{x \to - 1} \left( {{x^2} - x + 7} \right)$ bằng?
Tính $\mathop {\lim }\limits_{x \to - 2} \left( {3{x^2} - 3x - 8} \right)$ bằng?
Tính $\mathop {\lim }\limits_{x \to 2} \sqrt {\dfrac{{{x^4} + 3x - 1}}{{2{x^2} - 1}}} $ bằng?
Tính $\mathop {\lim }\limits_{x \to - \infty } \dfrac{{3{x^2} - 2x - 1}}{{{x^2} + 1}}$ bằng?
Tính $\mathop {\lim }\limits_{x \to {3^ + }} \dfrac{{\left| {x - 3} \right|}}{{3x - 9}}$ bằng?
Trong các mệnh đề sau đâu là mệnh đề đúng?
Tính $\mathop {\lim }\limits_{x \to 3} \dfrac{{{x^2} - 4x + 3}}{{{x^2} - 9}}$ bằng?
Tính $\mathop {\lim }\limits_{x \to - 1} \dfrac{{{x^2} + 6x + 5}}{{{x^3} + 2{x^2} - 1}}$ bằng?
Tính $\mathop {\lim }\limits_{x \to 2} \dfrac{{{x^3} - 6{x^2} + 11x - 6}}{{{x^2} - 4}}$ bằng?
Tính $\mathop {\lim }\limits_{x \to 3} \dfrac{{\sqrt {x + 1} - 2}}{{\sqrt {3x} - 3}}$ bằng?
Tính $\mathop {\lim }\limits_{x \to 2} \dfrac{{x - \sqrt {x + 2} }}{{\sqrt {4x + 1} - 3}}$ bằng?
Tính $\mathop {\lim }\limits_{x \to 0} \dfrac{{1 - \sqrt[3]{{x + 1}}}}{{3x}}$ bằng?
Tính$\mathop {\lim }\limits_{x \to - \infty } (x - 1)\sqrt {\dfrac{{{x^2}}}{{2{x^4} + {x^2} + 1}}} $ bằng?
Tính $\mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt {{x^2} + x + 3} - x} \right)$ bằng?
Tính $\mathop {\lim }\limits_{x \to - \infty } \left( {\sqrt {{x^2} + 1} + x - 1} \right)$ bằng?
Cho hàm số $f(x) = \sqrt {{x^2} + 2x + 4} - \sqrt {{x^2} - 2x + 4} $. Khẳng định nào sau đây là đúng?
Tính $\mathop {\lim }\limits_{x \to - \infty } \left( {\sqrt[3]{{{x^3} + 1}} + x - 1} \right)$ bằng?
Tính $\mathop {\lim }\limits_{x \to - \infty } x\sqrt {\dfrac{{3x + 2}}{{2{x^3} + {x^2} - 1}}} $ bằng?
Tính $\mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt[n]{{(x + 1)(x + 2)...(x + n)}} - x} \right)$ bằng:
Giới hạn \(\mathop {\lim }\limits_{x \to - \infty } \dfrac{{\sqrt {{x^2} + 3x + 5} }}{{4x - 1}}\).
