Giải bài 2 trang 75 sách bài tập toán 11 - Chân trời sáng tạo tập 1

Đề bài

Tìm các giới hạn sau:

a) \(\lim \frac{{2n - 3}}{{6n + 1}}\);

b) \(\lim \frac{{3n - 1}}{{{n^2} + n}}\);

c) \(\lim \frac{{\left( {2n - 1} \right)\left( {2n + 3} \right)}}{{2{n^2} + 4}}\);

d) \(\lim \frac{{4n + 1}}{{\sqrt {{n^2} + 3n}  + n}}\);

e) \(\lim \sqrt n \left( {\sqrt {n + 1}  - \sqrt n } \right)\);

g) \(\lim \frac{1}{{\sqrt {{n^2} + n}  - n}}\).

Lời giải chi tiết

a) \(\lim \frac{{2n - 3}}{{6n + 1}}\)\( = \lim \frac{{2 - \frac{3}{n}}}{{6 + \frac{1}{n}}}\)\( = \frac{{\lim 2 - \lim \frac{3}{n}}}{{\lim 6 + \lim \frac{1}{n}}}\)\( = \frac{{2 - 0}}{{6 + 0}} = \frac{1}{3}\);

b) \(\lim \frac{{3n - 1}}{{{n^2} + n}}\)\( = \lim \frac{{\frac{3}{n} - \frac{1}{{{n^2}}}}}{{1 + \frac{1}{n}}}\)\( = \frac{{\lim \frac{3}{n} - \lim \frac{1}{{{n^2}}}}}{{\lim 1 + \lim \frac{1}{n}}}\)\( = \frac{0}{{1 + 0}} = 0\);

c) \(\lim \frac{{\left( {2n - 1} \right)\left( {2n + 3} \right)}}{{2{n^2} + 4}}\)\( = \lim \frac{{\left( {2 - \frac{1}{n}} \right)\left( {2 + \frac{3}{n}} \right)}}{{2 + \frac{4}{{{n^2}}}}}\)\( = \frac{{\lim \left( {2 - \frac{1}{n}} \right)\lim \left( {2 + \frac{3}{n}} \right)}}{{\lim \left( {2 + \frac{4}{{{n^2}}}} \right)}}\)\( = \frac{{2.2}}{2} = 2\);

d) \(\lim \frac{{4n + 1}}{{\sqrt {{n^2} + 3n}  + n}}\)\( = \lim \frac{{4 + \frac{1}{n}}}{{\sqrt {1 + \frac{3}{n}}  + 1}}\)\( = \frac{{4 + \lim \frac{1}{n}}}{{\sqrt {1 + \lim \frac{3}{n}}  + 1}}\)\( = \frac{4}{{1 + 1}} = 2\);

e) \(\lim \sqrt n \left( {\sqrt {n + 1}  - \sqrt n } \right)\)\( = \lim \frac{{\sqrt n \left( {\sqrt {n + 1}  - \sqrt n } \right)\left( {\sqrt {n + 1}  + \sqrt n } \right)}}{{\sqrt {n + 1}  + \sqrt n }}\)\( = \lim \frac{{\sqrt n }}{{\sqrt {n + 1}  + \sqrt n }}\)

\( = \lim \frac{1}{{\sqrt {1 + \frac{1}{n}}  + \sqrt 1 }}\)\( = \frac{1}{{\sqrt {1 + \lim \frac{1}{n}}  + 1}} = \frac{1}{2}\)

g) \(\lim \frac{1}{{\sqrt {{n^2} + n}  - n}}\)\( = \lim \frac{{\sqrt {{n^2} + n}  + n}}{{\left( {\sqrt {{n^2} + n}  - n} \right)\left( {\sqrt {{n^2} + n}  + n} \right)}}\)\( = \lim \frac{{\sqrt {{n^2} + n}  + n}}{n}\)\( = \lim \frac{{\sqrt {1 + \frac{1}{n}}  + 1}}{1}\)\( = \frac{{\sqrt {1 + \lim \frac{1}{n}}  + 1}}{1} = 2\)