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