Đề bài
Tính các giới hạn sau:
a) \(\lim \frac{{6n - 5}}{{3n}}\)
b) \(\lim \frac{{ - 2{n^2} - 6n + 2}}{{8{n^2} - 5n + 4}}\)
c) \(\lim \frac{{{n^3} - 5n + 1}}{{3{n^2} - 4n + 2}}\)
d) \(\lim \frac{{ - 4n + 1}}{{9{n^2} - n + 2}}\)
e) \(\lim \frac{{\sqrt {4{n^2} + n + 1} }}{{8n + 3}}\)
g) \(\lim \frac{{{4^n} + {5^n}}}{{{{3.4}^n} - {{4.5}^n}}}\)
Phương pháp giải - Xem chi tiết
Sử dụng tính chất về dãy số có giới hạn vô cực và định lí về giới hạn hữu hạn.
Lời giải chi tiết
a) Ta có: \(\lim \frac{{6n - 5}}{{3n}} = \lim \frac{{n\left( {6 - \frac{5}{n}} \right)}}{{3n}} = \lim \frac{{6 - \frac{5}{n}}}{3} = \frac{{\lim 6 - \lim \frac{5}{n}}}{{\lim 3}} = \frac{6}{3} = 2\)
b) Ta có:
\(\lim \frac{{ - 2{n^2} - 6n + 2}}{{8{n^2} - 5n + 4}} = \lim \frac{{{n^2}\left( { - 2 - \frac{6}{n} + \frac{2}{{{n^2}}}} \right)}}{{{n^2}\left( {8 - \frac{5}{n} + \frac{4}{{{n^2}}}} \right)}} = \lim \frac{{ - 2 - \frac{6}{n} + \frac{2}{{{n^2}}}}}{{8 - \frac{5}{n} + \frac{4}{{{n^2}}}}}\)
\( = \frac{{\lim \left( { - 2} \right) - \lim \frac{6}{n} + \lim \frac{2}{{{n^2}}}}}{{\lim 8 - \lim \frac{5}{n} + \lim \frac{4}{{{n^2}}}}} = \frac{{ - 2}}{8} = \frac{{ - 1}}{4}\)
c) Ta có:
\(\lim \frac{{{n^3} - 5n + 1}}{{3{n^2} - 4n + 2}} = \lim \frac{{{n^3}\left( {1 - \frac{5}{{{n^2}}} + \frac{1}{{{n^3}}}} \right)}}{{{n^3}\left( {\frac{3}{n} - \frac{4}{{{n^2}}} + \frac{2}{{{n^3}}}} \right)}} = \lim \frac{{1 - \frac{5}{{{n^2}}} + \frac{1}{{{n^3}}}}}{{\frac{3}{n} - \frac{4}{{{n^2}}} + \frac{2}{{{n^3}}}}}\)
Vì \(\lim \left( {1 - \frac{5}{{{n^2}}} + \frac{1}{{{n^3}}}} \right) = \lim 1 - \lim \frac{5}{{{n^2}}} + \lim \frac{1}{{{n^3}}} = 1 - 0 + 0 = 1\),
Và \(\lim \left( {\frac{3}{n} - \frac{4}{{{n^2}}} + \frac{2}{{{n^3}}}} \right) = \lim \frac{3}{n} - \lim \frac{4}{{{n^2}}} + \lim \frac{2}{{{n^3}}} = 0\).
Ta suy ra:
\(\lim \frac{{{n^3} - 5n + 1}}{{3{n^2} - 4n + 2}} = \lim \frac{{1 - \frac{5}{{{n^2}}} + \frac{1}{{{n^3}}}}}{{\frac{3}{n} - \frac{4}{{{n^2}}} + \frac{2}{{{n^3}}}}} = + \infty \)
d) Ta có:
\(\begin{array}{l}\lim \frac{{ - 4n + 1}}{{9{n^2} - n + 2}} = \lim \frac{{{n^2}\left( {\frac{{ - 4}}{n} + \frac{1}{{{n^2}}}} \right)}}{{{n^2}\left( {9 - \frac{1}{n} + \frac{2}{{{n^2}}}} \right)}} = \lim \frac{{\frac{{ - 4}}{n} + \frac{1}{{{n^2}}}}}{{9 - \frac{1}{n} + \frac{2}{{{n^2}}}}} = \frac{{\lim \frac{{ - 4}}{n} + \lim \frac{1}{{{n^2}}}}}{{\lim 9 - \lim \frac{1}{n} + \lim \frac{2}{{{n^2}}}}}\\ = 0\end{array}\)
e) Ta có:
\(\lim \frac{{\sqrt {4{n^2} + n + 1} }}{{8n + 3}} = \lim \frac{{\sqrt {{n^2}\left( {4 + \frac{1}{n} + \frac{1}{{{n^2}}}} \right)} }}{{n\left( {8 + \frac{3}{n}} \right)}} = \lim \frac{{n\sqrt {4 + \frac{1}{n} + \frac{1}{{{n^2}}}} }}{{n\left( {8 + \frac{3}{n}} \right)}}\)
\( = \lim \frac{{\sqrt {4 + \frac{1}{n} + \frac{1}{{{n^2}}}} }}{{8 + \frac{3}{n}}} = \frac{{\sqrt {\lim 4 + \lim \frac{1}{n} + \lim \frac{1}{{{n^2}}}} }}{{\lim 8 + \lim \frac{3}{n}}} = \frac{{\sqrt 4 }}{8} = \frac{2}{8} = \frac{1}{4}\)
f) Ta có:
\(\lim \frac{{{4^n} + {5^n}}}{{{{3.4}^n} - {{4.5}^n}}} = \lim \frac{{\frac{{{4^n}}}{{{5^n}}} + 1}}{{3.\frac{{{4^n}}}{{{5^n}}} - 4}} = \frac{{\lim {{\left( {\frac{4}{5}} \right)}^n} + \lim 1}}{{\lim 3.\lim {{\left( {\frac{4}{5}} \right)}^n} - \lim 4}} = \frac{{0 + 1}}{{3.0 - 4}} =- \frac{1}{4}\)
