Tính \(\mathop {\lim }\limits_{x \to  + \infty } \...

B

- Hướng dẫn giải

Phương pháp giải:

- Đặt \(x = \frac{1}{y}\), khi \(x \to  + \infty :\,\,\,y \to 0\) .

- Nhân liên hợp, tính \(\mathop {\lim }\limits_{y \to 0} \frac{{\sqrt[n]{{(1 + y)(1 + 2y)...(1 + ny)}} - 1}}{y}\).

Giải chi tiết:

Đặt \(x = \frac{1}{y}\), khi \(x \to  + \infty :\,\,\,y \to 0\)

\(\mathop {\lim }\limits_{x \to  + \infty } \left( {\sqrt[n]{{(x + 1)(x + 2)...(x + n)}} - x} \right) \)

\(= \mathop {\lim }\limits_{y \to 0} \left( {\sqrt[n]{{\left( {\frac{1}{y} + 1} \right)\left( {\frac{1}{y} + 2} \right)...\left( {\frac{1}{y} + n} \right)}} - \frac{1}{y}} \right)\)

\( = \mathop {\lim }\limits_{y \to 0} \frac{{\sqrt[n]{{(1 + y)(1 + 2y)...(1 + ny)}} - 1}}{y}\)

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

Tổng quát:

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

Khi đó:

\(\mathop {\lim }\limits_{y \to 0} \frac{{\sqrt[n]{{(1 + y)(1 + 2y)...(1 + ny)}} - 1}}{y} = \frac{1}{n} + \frac{2}{n} + \frac{3}{n} + ... + \frac{n}{n} = \frac{{1 + 2 + 3 + ... + n}}{n} = \frac{{\frac{{n(n + 1)}}{2}}}{n} = \frac{{n + 1}}{2}\)

Chọn: B.