Giải bài 42 trang 83 sách bài tập toán 11 – Cánh diều

Đề bài

Tính các giới hạn sau:

a) \(\lim \frac{{2n – 4}}{5}\)                      

b) \(\lim \frac{{1 + \frac{1}{{2n}}}}{{2n}}\)                

c) \(\lim \left( {2 + \frac{7}{{{4^n}}}} \right)\)

d) \(\lim \frac{{ – 4{n^2} – 3}}{{2{n^2} – n + 5}}\)               

e) \(\lim \frac{{\sqrt {9{n^2} + 2n + 1} }}{{n – 5}}\)     

g) \(\lim \frac{{{3^n} + {{4.9}^n}}}{{{{3.4}^n} + {9^n}}}\)

Lời giải chi tiết

a) Ta có \(\lim \left( {2n – 4} \right) = \lim \left[ {n\left( {2 – \frac{4}{n}} \right)} \right] = \lim n.\lim \left( {2 – \frac{4}{n}} \right) = 2\lim n =  + \infty \)

Suy ra \(\lim \frac{{2n – 4}}{5} =  + \infty \).

b) Ta có: \(\lim \left( {1 + \frac{1}{{2n}}} \right) = \lim 1 + \lim \frac{1}{{2n}} = 1 + 0 = 1\) và \(\lim 2n =  + \infty \).

Suy ra \(\lim \frac{{1 + \frac{1}{{2n}}}}{{2n}} = 0\).

c) Ta có \(\lim \left( {2 + \frac{7}{{{4^n}}}} \right) = \lim 2 + \lim \frac{7}{{{4^n}}} = 2 + 0 = 2\).

d) Ta có \(\lim \frac{{ – 4{n^2} – 3}}{{2{n^2} – n + 5}} = \lim \frac{{{n^2}\left( { – 4 – \frac{3}{{{n^2}}}} \right)}}{{{n^2}\left( {2 – \frac{1}{n} + \frac{5}{{{n^2}}}} \right)}}\)

\( = \lim \frac{{ – 4 – \frac{3}{{{n^2}}}}}{{2 – \frac{1}{n} + \frac{5}{{{n^2}}}}} = \frac{{\lim \left( { – 4} \right) – \lim \frac{3}{{{n^2}}}}}{{\lim 2 – \lim \frac{1}{n} + \lim \frac{5}{{{n^2}}}}} = \frac{{ – 4 – 0}}{{2 – 0 + 0}} =  – 2\)

e) Ta có: \(\lim \frac{{\sqrt {9{n^2} + 2n + 1} }}{{n – 5}} = \lim \frac{{\sqrt {{n^2}\left( {9 + \frac{2}{n} + \frac{1}{{{n^2}}}} \right)} }}{{n\left( {1 – \frac{5}{n}} \right)}} = \lim \frac{{\sqrt {{n^2}} \sqrt {9 + \frac{2}{n} + \frac{1}{{{n^2}}}} }}{{n\left( {1 – \frac{5}{n}} \right)}}\)

\( = \lim \frac{{n\sqrt {9 + \frac{2}{n} + \frac{1}{{{n^2}}}} }}{{n\left( {1 – \frac{5}{n}} \right)}} = \lim \frac{{\sqrt {9 + \frac{2}{n} + \frac{1}{{{n^2}}}} }}{{1 – \frac{5}{n}}}\).

Do \(\lim \left( {9 + \frac{2}{n} + \frac{1}{{{n^2}}}} \right) = \lim 9 + \lim \frac{2}{n} + \lim \frac{1}{{{n^2}}} = 9 + 0 + 0 = 9\), ta suy ra:

\(\lim \sqrt {9 + \frac{2}{n} + \frac{1}{{{n^2}}}}  = \sqrt 9  = 3\).

Mặt khác, \(\lim \left( {1 – \frac{5}{n}} \right) = \lim 1 – \lim \frac{5}{n} = 1 – 0 = 1\)

Suy ra \(\lim \frac{{\sqrt {9{n^2} + 2n + 1} }}{{n – 5}} = \lim \frac{{\sqrt {9 + \frac{2}{n} + \frac{1}{{{n^2}}}} }}{{1 – \frac{5}{n}}} = \frac{3}{1} = 3\).

f) Ta có: \(\lim \frac{{{3^n} + {{4.9}^n}}}{{{{3.4}^n} + {9^n}}} = \lim \frac{{\frac{{{3^n}}}{{{9^n}}} + 4}}{{3.\frac{{{4^n}}}{{{9^n}}} + 1}} = \frac{{\lim {{\left( {\frac{3}{9}} \right)}^n} + \lim 4}}{{\lim 3.\lim {{\left( {\frac{4}{9}} \right)}^n} + \lim 1}} = \frac{{0 + 4}}{{3.0 + 1}} = 4\)