Giải bài 9 trang 69 sách bài tập toán 11 – Cánh diều

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) \(\lim \frac{{{4^n} + {5^n}}}{{{{3.4}^n} – {{4.5}^n}}}\)

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:

\(\lim \frac{{{4^n} + {5^n}}}{{{{3.4}^n} – {{4.5}^n}}} = \lim \frac{{\frac{{{4^n}}}{{{5^n}}} + 1}}{{3.\frac{{{4^n}}}{{{5^n}}} – 4}} = \frac{{\lim {{\left( {\frac{4}{5}} \right)}^n} + \lim 1}}{{\lim 3.\lim {{\left( {\frac{4}{5}} \right)}^n} – \lim 4}} = \frac{{0 + 1}}{{3.0 – 4}} = \frac{1}{4}\)