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