- LG a
- LG b
Tính giới hạn \[\mathop {\lim }\limits_{n \to + \infty } {x_n}\]
LG a
\[{x_n} = \frac{{\sqrt {{n^2} + 1} + \sqrt n }}{{\sqrt[3]{{{n^3} + n}} - n}}\]
Lời giải chi tiết:
\[\begin{array}{l}\lim {x_n} = \lim \frac{{\sqrt {{n^2} + 1} + \sqrt n }}{{\sqrt[3]{{{n^3} + n}} - n}}\\ = \lim \frac{{\sqrt {{n^2}\left[ {1 + \frac{1}{{{n^2}}}} \right]} + \frac{n}{{\sqrt n }}}}{{\sqrt[3]{{{n^3}\left[ {1 + \frac{1}{{{n^2}}}} \right]}} - n}}\\ = \lim \frac{{n\sqrt {1 + \frac{1}{{{n^2}}}} + n.\frac{1}{{\sqrt n }}}}{{n\sqrt[3]{{1 + \frac{1}{{{n^2}}}}} - n}}\\ = \lim \frac{{n\left[ {\sqrt {1 + \frac{1}{{{n^2}}}} + \frac{1}{{\sqrt n }}} \right]}}{{n\left[ {\sqrt[3]{{1 + \frac{1}{{{n^2}}}}} - 1} \right]}}\\ = \lim \frac{{\sqrt {1 + \frac{1}{{{n^2}}}} + \frac{1}{{\sqrt n }}}}{{\sqrt[3]{{1 + \frac{1}{{{n^2}}}}} - 1}} = + \infty \end{array}\]
Vì \[\lim \left[ {\sqrt {1 + \frac{1}{{{n^2}}}} + \frac{1}{{\sqrt n }}} \right] = 1 > 0\] và \[\left\{ \begin{array}{l}\lim \left[ {\sqrt[3]{{1 + \frac{1}{{{n^2}}}}} - 1} \right] = 0\\\sqrt[3]{{1 + \frac{1}{{{n^2}}}}} - 1 > 0\end{array} \right.\]
LG b
\[{x_n} = \left[ {n - \frac{1}{n}} \right]\left[ {\frac{{1 - 4n}}{{2{n^2}}}} \right]\]
Lời giải chi tiết:
\[\begin{array}{l}\lim {x_n} = \lim \left[ {n - \frac{1}{n}} \right]\left[ {\frac{{1 - 4n}}{{2{n^2}}}} \right]\\ = \lim \frac{{\left[ {{n^2} - 1} \right]\left[ {1 - 4n} \right]}}{{2{n^3}}}\\ = \lim \frac{{\frac{{{n^2} - 1}}{{{n^2}}}.\frac{{1 - 4n}}{n}}}{{\frac{{2{n^3}}}{{{n^3}}}}}\\ = \lim \frac{{\left[ {1 - \frac{1}{{{n^2}}}} \right].\left[ {\frac{1}{n} - 4} \right]}}{2}\\ = \frac{{\left[ {1 - 0} \right]\left[ {0 - 4} \right]}}{2} = - 2\end{array}\]
Cách khác:
\[\begin{array}{l}\lim {x_n} = \lim \left[ {n - \frac{1}{n}} \right]\left[ {\frac{{1 - 4n}}{{2{n^2}}}} \right]\\ = \lim \left[ {n\left[ {1 - \frac{1}{{{n^2}}}} \right].\frac{1}{n}\left[ {\frac{{1 - 4n}}{{2n}}} \right]} \right]\\ = \lim \left[ {n\left[ {1 - \frac{1}{{{n^2}}}} \right].\frac{1}{n}\left[ {\frac{1}{{2n}} - 2} \right]} \right]\\ = \lim \left[ {1 - \frac{1}{{{n^2}}}} \right]\left[ {\frac{1}{{2n}} - 2} \right]\\ = \left[ {1 - 0} \right]\left[ {0 - 2} \right] = - 2\end{array}\]