Bài 3.1 trang 64 SGK Toán 11 tập 1 - Cùng khám phá

Đề bài

Tìm các giới hạn:

a, \(\lim \frac{{3n + 2}}{{4 - n}}\)

b, \(\lim \frac{{5{n^2} + 2n - 1}}{{2{n^2} + n + 1}}\)

c, \(\lim \frac{{\sqrt {{n^2} + 4n + 2} }}{{3n - 1}}\)

d, \(\lim \frac{{n + 7}}{{4 + {n^2}}}\)

e, \(\lim \frac{{{2^n} - 1}}{{{5^n} + 1}}\)

Lời giải chi tiết

a, Ta có: \(\frac{{3n + 2}}{{4 - n}} = \frac{{3 + \frac{2}{n}}}{{\frac{4}{n} - 1}}\)

Vì lim 3= 3, lim \(\frac{2}{n}\)=0, lim\(\frac{4}{n}\)=0, lim 1=1 nên \(\lim (3 + \frac{2}{n}) = 3\) và \(\lim (\frac{4}{n} - 1)\)= -1

Vậy \(\lim \frac{{3n + 2}}{{4 - n}} = - 3\).

b, Ta có: \(\frac{{5{n^2} + 2n - 1}}{{2{n^2} + n + 1}} = \frac{{5 + \frac{2}{n} - \frac{1}{{{n^2}}}}}{{2 + \frac{1}{n} + \frac{1}{{{n^2}}}}}\)

Vì lim 5= 5, lim 2=2, \(\lim \frac{2}{n} = 0\), \(\lim \frac{1}{n} = 0\), \(\lim \frac{1}{{{n^2}}} = 0\) nên \(\lim (5 + \frac{2}{n} - \frac{1}{{{n^2}}}) = 5\) và \(\lim (2 + \frac{1}{n} + \frac{1}{{{n^2}}}) = 2\).

Vậy \(\lim \frac{{5{n^2} + 2n - 1}}{{2{n^2} + n + 1}} = \frac{5}{2}\).

c, Ta có: \(\)\(\frac{{\sqrt {{n^2} + 4n + 2} }}{{3n - 1}} = \frac{{\frac{{\sqrt {{n^2} + 4n + 2} }}{n}}}{{\frac{{3n - 1}}{n}}} = \frac{{\sqrt {\frac{{{n^2} + 4n + 2}}{{{n^2}}}} }}{{3 - \frac{1}{n}}}\)=\(\frac{{\sqrt {1 + \frac{4}{n} + \frac{2}{{{n^2}}}} }}{{3 - \frac{1}{n}}}\)

Vì lim 1=1, lim 3=3, \(\lim \frac{4}{n} = 0\), \(\lim \frac{2}{{{n^2}}} = 0\), \(\lim \frac{1}{n} = 0\) nên \(\lim \sqrt {1 + \frac{4}{n} + \frac{2}{{{n^2}}}} = \lim \sqrt 1 = 1\) và \(\lim (3 - \frac{1}{n}) = 3\)

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

d, Ta có: \(\frac{{n + 7}}{{4 + {n^2}}} = \frac{{\frac{1}{n} + \frac{7}{{{n^2}}}}}{{\frac{4}{{{n^2}}} + 1}}\)

Vì lim 1=1, \(\lim \frac{1}{n} = 0\); \(\lim \frac{7}{{{n^2}}} = 0\); \(\lim \frac{4}{{{n^2}}} = 0\) nên \(\lim (\frac{1}{n} + \frac{7}{{{n^2}}}) = 0\) và \(\lim (\frac{4}{{{n^2}}} + 1) = 1\)

Vậy \(\lim \frac{{n + 7}}{{4 + {n^2}}} = 0\).

e, Ta có: \(\frac{{{2^n} - 1}}{{{5^n} + 1}} = \frac{{{{(\frac{2}{5})}^n} - \frac{1}{{{5^n}}}}}{{1 + \frac{1}{{{5^n}}}}}\)

Vì lim 1=1 , \(\lim {(\frac{2}{5})^n} = 0\), \(\lim \frac{1}{{{5^n}}} = 0\) nên \(\lim \left[ {{{\left( {\frac{2}{5}} \right)}^n} - \frac{1}{{{5^n}}}} \right] = 0\) và \(\lim \left( {1 + \frac{1}{{{5^n}}}} \right) = 1\)

Vậy \(\lim \frac{{{2^n} - 1}}{{{5^n} + 1}} = 0\).