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, \(\lim \frac{{{2^n} - 1}}{{{5^n} + 1}}\)
Áp dụng tính chất: \(\lim \frac{1}{n} = 0\),
\(\lim \frac{1}{{{n^k}}} = 0\) với k là số nguyên dương;
\(\lim {q^n} = 0\)( nếu \(\left| q \right|
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
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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 \(\lim \frac{{{2^n} - 1}}{{{5^n} + 1}} = 0\).