Tính $\mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt[n]{{(x + 1)(x + 2)...(x + n)}} - x} \right)$ bằng:
-
A.
$0$
-
B.
$\dfrac{{n + 1}}{2}$
-
C.
$n$
-
D.
$1$
- Đặt $x = \dfrac{1}{y}$, khi $x \to + \infty :\,\,\,y \to 0$ .
- Nhân liên hợp, tính $\mathop {\lim }\limits_{y \to 0} \dfrac{{\sqrt[n]{{(1 + y)(1 + 2y)...(1 + ny)}} - 1}}{y}$.
Đặt $x = \dfrac{1}{y}$, khi $x \to + \infty :\,\,\,y \to 0$
$\mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt[n]{{(x + 1)(x + 2)...(x + n)}} - x} \right) = \mathop {\lim }\limits_{y \to 0} \left( {\sqrt[n]{{\left( {\dfrac{1}{y} + 1} \right)\left( {\dfrac{1}{y} + 2} \right)...\left( {\dfrac{1}{y} + n} \right)}} - \dfrac{1}{y}} \right) = \mathop {\lim }\limits_{y \to 0} \dfrac{{\sqrt[n]{{(1 + y)(1 + 2y)...(1 + ny)}} - 1}}{y}$
$\begin{array}{l}\sqrt[n]{{(1 + y)(1 + 2y)...(1 + ny)}} - 1\\ = \sqrt[n]{{1 + y}} - \sqrt[n]{{1 + y}} + \sqrt[n]{{\left( {1 + y} \right)\left( {1 + 2y} \right)}} - \sqrt[n]{{\left( {1 + y} \right)\left( {1 + 2y} \right)}} + ... + \sqrt[n]{{(1 + y)(1 + 2y)...(1 + (n - 1)y)}}\\\,\,\,\,\, - \sqrt[n]{{(1 + y)(1 + 2y)...(1 + (n - 1)y)}} + \sqrt[n]{{(1 + y)(1 + 2y)...(1 + ny)}} - 1\\ = \left( {\sqrt[n]{{1 + y}} - 1} \right) + \sqrt[n]{{1 + y}}\left( {\sqrt[n]{{1 + 2y}} - 1} \right) + ... + \sqrt[n]{{(1 + y)(1 + 2y)...(1 + (n - 1)y)}}\left( {\sqrt[n]{{1 + ny}} - 1} \right)\\ \Rightarrow \mathop {\lim }\limits_{y \to 0} \dfrac{{\sqrt[n]{{(1 + y)(1 + 2y)...(1 + ny)}} - 1}}{y} = \mathop {\lim }\limits_{y \to 0} \left[ {\dfrac{{\left( {\sqrt[n]{{1 + y}} - 1} \right)}}{y}} \right] + \mathop {\lim }\limits_{y \to 0} \left[ {\sqrt[n]{{1 + y}}.\dfrac{{\left( {\sqrt[n]{{1 + 2y}} - 1} \right)}}{y}} \right] + ... + \\\,\,\,\,\,\mathop {\lim }\limits_{y \to 0} \left[ {\sqrt[n]{{(1 + y)(1 + 2y)...(1 + (n - 1)y)}}.\dfrac{{\left( {\sqrt[n]{{1 + ny}} - 1} \right)}}{y}} \right]\end{array}$
Tổng quát:
$\begin{array}{l}\mathop {\lim }\limits_{y \to 0} \left[ {\sqrt[n]{{(1 + y)(1 + 2y)...(1 + (k - 1)y)}}.\dfrac{{\sqrt[n]{{1 + ky}} - 1}}{y}} \right]\\ = \mathop {\lim }\limits_{y \to 0} \left[ {\sqrt[n]{{(1 + y)(1 + 2y)...(1 + (k - 1)y)}}.\dfrac{{\left( {\sqrt[n]{{1 + ky}} - 1} \right)\left[ {{{\left( {\sqrt[n]{{1 + ky}}} \right)}^{n - 1}} + {{\left( {\sqrt[n]{{1 + ky}}} \right)}^{n - 2}} + ... + 1} \right]}}{{y\left[ {{{\left( {\sqrt[n]{{1 + ky}}} \right)}^{n - 1}} + {{\left( {\sqrt[n]{{1 + ky}}} \right)}^{n - 2}} + ... + 1} \right]}}} \right]\\ = \mathop {\lim }\limits_{y \to 0} \dfrac{{(1 + ky - 1).\sqrt[n]{{(1 + y)(1 + 2y)...(1 + (k - 1)y)}}}}{{y{{\left( {\sqrt[n]{{1 + ky}}} \right)}^{n - 1}} + {{\left( {\sqrt[n]{{1 + ky}}} \right)}^{n - 2}} + ... + 1}}\\ = \mathop {\lim }\limits_{y \to 0} \dfrac{{k.\sqrt[n]{{(1 + y)(1 + 2y)...(1 + (k - 1)y)}}}}{{{{\left( {\sqrt[n]{{1 + ky}}} \right)}^{n - 1}} + {{\left( {\sqrt[n]{{1 + ky}}} \right)}^{n - 2}} + ... + 1}} = \dfrac{k}{n}\end{array}$
Khi đó:
$\mathop {\lim }\limits_{y \to 0} \dfrac{{\sqrt[n]{{(1 + y)(1 + 2y)...(1 + ny)}} - 1}}{y} = \dfrac{1}{n} + \dfrac{2}{n} + \dfrac{3}{n} + ... + \dfrac{n}{n} = \dfrac{{1 + 2 + 3 + ... + n}}{n} = \dfrac{{\dfrac{{n(n + 1)}}{2}}}{n} = \dfrac{{n + 1}}{2}$
Đáp án : B
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