Tính $\underset{x\to +\infty }{\lim }\left(\sqrt[n]{\left(x+1\right)\left(x+2\right)\mathrm{...}\left(x+n\right)}-x\right)$ bằng

Đặt $x=\frac{1}{y}$ khi $x\to +\infty :y\to 0$

$\begin{array}{l}\underset{x\to +\infty }{\lim }\left(\sqrt[n]{\left(x+1\right)\left(x+2\right)\mathrm{...}\left(x+n\right)}-x\right)\\ =\underset{x\to 0}{\lim }\left(\sqrt[n]{\left(\frac{1}{y}+1\right)\left(\frac{1}{y}+2\right)\mathrm{...}\left(\frac{1}{y}+n\right)}-\frac{1}{y}\right)\\ =\underset{x\to 0}{\lim }\frac{\sqrt[n]{(1+y)(1+2y)\mathrm{...}(1+ny)}-1}{y}\end{array}$

$\begin{array}{l}*\sqrt[n]{(1+y)(1+2y)\mathrm{...}(1+ny)}-1\\ =\sqrt[n]{1+y}-\sqrt[n]{1+y}+\sqrt[n]{(1+y)(1+2y)}-\sqrt[n]{(1+y)(1+2y)+\mathrm{...}}\\ -\sqrt[n]{(1+y)(1+2y)\mathrm{...}(1+(n-1\left)y\right)}+\sqrt[n]{(1+y)(1+2y)\mathrm{...}(1+ny)}-1\\ =\left(\sqrt[n]{1+y}-1\right)+\sqrt[n]{1+y}\left(\sqrt[n]{1+2y}-1\right)+\mathrm{...}+\sqrt[n]{(1+y)(1+2y)\mathrm{...}(1+(n-1\left)y\right)}\left(\sqrt[n]{1+ny}-1\right)\\ \Rightarrow \underset{y\to 0}{\lim }\frac{\sqrt[n]{(1+y)(1+2y)\mathrm{...}(1+ny)}-1}{y}\\ =\underset{y\to 0}{\lim }\left[\frac{\left(\sqrt[n]{1+y}-1\right)}{y}\right]+\underset{y\to 0}{\lim }\left[\sqrt[n]{1+y}\frac{\left(\sqrt[n]{1+2y}-1\right)}{y}\right]+\mathrm{...}\\ +\underset{y\to 0}{\lim }\left[\sqrt[n]{(1+y)(1+2y)\mathrm{...}(1+(n-1\left)y\right)}.\frac{\left(\sqrt[n]{1+ny}-1\right)}{y}\right]\end{array}$

Tổng quát

$\begin{array}{l}\underset{y\to 0}{\lim }\left[\sqrt[n]{(1+y)(1+2y)\mathrm{...}(1+(k-1)y}.\frac{\sqrt[n]{1+ky}-1}{y}\right]\\ =\underset{y\to 0}{\lim }\left[\sqrt[n]{(1+y)(1+2y)\mathrm{...}(1+(k-1)y}.\frac{\left(\sqrt[n]{1+ky}-1\right)\left[{\left(\sqrt[n]{1+ky}\right)}^{n-1}+{\left(\sqrt[n]{1+ky}\right)}^{n-2}+\mathrm{...}+1\right]}{y\left[{\left(\sqrt[n]{1+ky}\right)}^{n-1}+{\left(\sqrt[n]{1+ky}\right)}^{n-2}+\mathrm{...}+1\right]}\right]\\ =\underset{y\to 0}{\lim }\frac{(1+ky-1).\sqrt[n]{(1+y)(1+2y)\mathrm{...}(1+(k-1\left)y\right)}}{{\left(\sqrt[n]{1+ky}\right)}^{n-1}+{\left(\sqrt[n]{1+ky}\right)}^{n-2}+\mathrm{...}+1}=\frac{k}{n}\end{array}$

Khi đó

$\begin{array}{l}\underset{y\to 0}{\lim }\frac{\sqrt[n]{(1+y)(1+2y)\mathrm{...}(1+ny)}-1}{y}=\frac{1}{n}+\frac{2}{n}+\frac{3}{n}+\mathrm{...}+\frac{n}{n}\\ =\frac{1+2+3+\mathrm{...}+n}{n}=\frac{\frac{n(n+1)}{2}}{n}=\frac{n+1}{2}\end{array}$

Đáp án cần chọn là: B