Hướng dẫn giải (chi tiết)
Đặt $x=\dfrac{1}{y}$, khi $x \rightarrow+\infty : y \rightarrow 0$
Ta có $f\left( y \right) = \sqrt[n]{{\left( {\dfrac{1}{y} + 1} \right)\left( {\dfrac{1}{y} + 2} \right) \ldots \left( {\dfrac{1}{y} + n} \right)}} - \dfrac{1}{y}$ và $\mathop {\lim }\limits_{x \to + \infty } f\left( x \right) = \mathop {\lim }\limits_{y \to 0} f\left( y \right)$
Suy ra
$\begin{array}{*{20}{l}}\n{\mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt[n]{{(x + 1)(x + 2) \ldots (x + n)}} - x} \right) = \mathop {\lim }\limits_{y \to 0} \left( {\sqrt[n]{{\left( {\dfrac{1}{y} + 1} \right)\left( {\dfrac{1}{y} + 2} \right) \ldots \left( {\dfrac{1}{y} + n} \right)}} - \dfrac{1}{y}} \right)}\\\n{ = \mathop {\lim }\limits_{y \to 0} \dfrac{{\sqrt[n]{{(1 + y)(1 + 2y) \ldots (1 + ny)}} - 1}}{y}}\n\end{array}$
$\sqrt[n]{(1+y)(1+2 y) \ldots(1+n y)}-1$
$=\sqrt[n]{1+y}-\sqrt[n]{1+y}+\sqrt[n]{(1+y)(1+2 y)}-\sqrt[n]{(1+y)(1+2 y)}+\ldots$
$+\sqrt[n]{(1+y)(1+2 y) \ldots(1+(n-1) y)}$
$-\sqrt[n]{(1+y)(1+2 y) \ldots(1+(n-1) y)}+\sqrt[n]{(1+y)(1+2 y) \ldots(1+n y)}-1$
$=(\sqrt[n]{1+y}-1)+\sqrt[n]{1+y}(\sqrt[n]{1+2 y}-1)+\ldots$
$+\sqrt[n]{(1+y)(1+2 y) \ldots(1+(n-1) y)}(\sqrt[n]{1+n y}-1)$
$\Rightarrow\mathop {\lim }\limits_{y \to 0} \dfrac{\sqrt[5]{(1+y)(1+2 y) \cdots(1+n y)}-1}{y}\\=\mathop {\lim }\limits_{y \to 0} \left[\dfrac{(\sqrt[n]{1+y}-1)}{y}\right]+\mathop {\lim }\limits_{y \to 0} \left[\sqrt[n]{1+y} \cdot \dfrac{(\sqrt[n]{1+2 y}-1)}{y}\right]+\ldots+\mathop {\lim }\limits_{y \to 0} \left[\sqrt[n]{(1+y)(1+2 y) \ldots(1+(n-1) y)} \cdot \dfrac{(\sqrt[n]{1+n y}-1)}{y}\right]$
Tổng quát:
$\mathop {\lim }\limits_{y \to 0} \left[\sqrt[n]{(1+y)(1+2 y) \ldots(1+(k-1) y)} \cdot \dfrac{\sqrt[n]{1+k y}-1}{y}\right]$
$=\mathop {\lim }\limits_{y \to 0} {\rm{ }}\left( {\sqrt[n]{{(1 + y)(1 + 2y) \ldots (1 + (k - 1)y)}}.\dfrac{{(\sqrt[n]{{1 + ky}} - 1)\left[ {{{(\sqrt[n]{{1 + ky}})}^{n - 1}} + {{(\sqrt[n]{{1 + ky}})}^{n - 2}} + \ldots + 1} \right]}}{{y\left[ {{{(\sqrt[n]{{1 + ky}})}^{n - 1}} + {{(\sqrt[n]{{1 + ky}})}^{n - 2}} + \ldots + 1} \right]}}} \right)$
$=\mathop {\lim }\limits_{y \to 0} \dfrac{(1+k y-1) \cdot \sqrt[n]{(1+y)(1+2 y) \ldots(1+(k-1) y)}}{y(\sqrt[n]{1+k y})^{n-1}+(\sqrt[n]{1+k y})^{n-2}+\ldots+1}$
$=\mathop {\lim }\limits_{y \to 0} \dfrac{k \cdot \sqrt[n]{(1+y)(1+2 y) \ldots(1+(k-1) y)}}{(\sqrt[n]{1+k y})^{n-1}+(\sqrt[n]{1+k y})^{n-2}+\ldots+1}=\dfrac{k}{n}$
Khi đó:
$\begin{array}{l}{\mathop {\lim }\limits_{y \to 0} \dfrac{\sqrt[n]{(1+y)(1+2 y) \ldots(1+n y)}-1}{y}=\dfrac{1}{n}+\dfrac{2}{n}+\dfrac{3}{n}+\ldots+\dfrac{n}{n}=\dfrac{1+2+3+\ldots+n}{n}} \\ {=\dfrac{\dfrac{n(n+1)}{2}}{n}=\dfrac{n+1}{2}}\end{array}$
Đáp án cần chọn là: B.
