Bài 2 trang 79 SGK Toán 11 tập 1 – Cánh Diều

Đề bài

Tính các giới hạn sau:

a) \(\lim \frac{{2{n^2} + 6n + 1}}{{8{n^2} + 5}}\)                            

b) \(\lim \frac{{4{n^2} – 3n + 1}}{{ – 3{n^3} + 5{n^2} – 2}}\);          

c) \(\lim \frac{{\sqrt {4{n^2} – n + 3} }}{{8n – 5}}\);

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

e) \(\lim \frac{{{{4.5}^n} + {2^{n + 2}}}}{{{{6.5}^n}}}\)                         

g) \(\lim \frac{{2 + \frac{4}{{{n^3}}}}}{{{6^n}}}\).

Lời giải chi tiết

a) \(\lim \frac{{2{n^2} + 6n + 1}}{{8{n^2} + 5}} = \lim \frac{{{n^2}\left( {2 + \frac{6}{n} + \frac{1}{{{n^2}}}} \right)}}{{{n^2}\left( {8 + \frac{5}{{{n^2}}}} \right)}} = \lim \frac{{2 + \frac{6}{n} + \frac{1}{n}}}{{8 + \frac{5}{n}}} = \frac{2}{8} = \frac{1}{4}\)

b) \(\lim \frac{{4{n^2} – 3n + 1}}{{ – 3{n^3} + 6{n^2} – 2}} = \lim \frac{{{n^3}\left( {\frac{4}{n} – \frac{3}{{{n^2}}} + \frac{1}{{{n^3}}}} \right)}}{{{n^3}\left( { – 3 + \frac{6}{n} – \frac{2}{{{n^3}}}} \right)}} = \lim \frac{{\frac{4}{n} – \frac{3}{{{n^2}}} + \frac{1}{{{n^3}}}}}{{ – 3 + \frac{6}{n} – \frac{2}{{{n^3}}}}} = \frac{{0 – 0 + 0}}{{ – 3 + 0 – 0}} = 0\).

c) \(\lim \frac{{\sqrt {4{n^2} – n + 3} }}{{8n – 5}} = \lim \frac{{n\sqrt {4 – \frac{1}{n} + \frac{3}{{{n^2}}}} }}{{n\left( {8 – \frac{5}{n}} \right)}} = \frac{{\sqrt {4 – 0 + 0} }}{{8 – 0}} = \frac{2}{8} = \frac{1}{4}\).

d) \(\lim \left( {4 – \frac{{{2^{{\rm{n}} + 1}}}}{{{3^{\rm{n}}}}}} \right) = \lim \left( {4 – 2 \cdot {{\left( {\frac{2}{3}} \right)}^{\rm{n}}}} \right) = 4 – 2.0 = 4\).

e) \(\lim \frac{{{{4.5}^{\rm{n}}} + {2^{{\rm{n}} + 2}}}}{{{{6.5}^{\rm{n}}}}} = \lim \frac{{{{4.5}^{\rm{n}}} + {2^2}{{.2}^{\rm{n}}}}}{{{{6.5}^{\rm{n}}}}} = \lim \frac{{{5^n}.\left[ {4 + 4.{{\left( {\frac{2}{5}} \right)}^{\rm{n}}}} \right]}}{{{{6.5}^n}}} = \lim \frac{{4 + 4.{{\left( {\frac{2}{5}} \right)}^{\rm{n}}}}}{6} = \frac{{4 + 4.0}}{6} = \frac{2}{3}\).

g) \(\lim \frac{{2 + \frac{4}{{{n^3}}}}}{{{6^{\rm{n}}}}} = \lim \left( {2 + \frac{4}{{{{\rm{n}}^3}}}} \right).\lim {\left( {\frac{1}{6}} \right)^{\rm{n}}} = \left( {2 + 0} \right).0 = 0\).