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Bài 16. Tìm các giới hạn sau :
a. \(\lim {{{n^2} + 4n – 5} \over {3{n^3} + {n^2} + 7}}\)
b. \(\lim {{{n^5} + {n^4} – 3n – 2} \over {4{n^3} + 6{n^2} + 9}}\)
c. \(\lim {{\sqrt {2{n^4} + 3n – 2} } \over {2{n^2} – n + 3}}\)
d. \(\lim {{{3^n} – {{2.5}^n}} \over {7 + {{3.5}^n}}}\)
a.
\(\eqalign{
& \lim {{{n^2} + 4n – 5} \over {3{n^3} + {n^2} + 7}} = \lim {{{n^3}\left( {{1 \over n} + {4 \over {{n^2}}} – {5 \over {{n^3}}}} \right)} \over {{n^3}\left( {3 + {1 \over n} + {7 \over {{n^3}}}} \right)}} \cr
& = \lim {{{1 \over n} + {4 \over {{n^2}}} – {5 \over {{n^3}}}} \over {3 + {1 \over n} + {7 \over {{n^3}}}}} = {0 \over 3} = 0 \cr} \)
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b.
\(\eqalign{
& \lim {{{n^5} + {n^4} – 3n – 2} \over {4{n^3} + 6{n^2} + 9}} = \lim {n^2}{{{n^3}\left( {1 + {1 \over n} – {3 \over {{n^4}}} – {2 \over {{n^5}}}} \right)} \over {{n^3}\left( {4 + {6 \over n} + {9 \over {{n^3}}}} \right)}} \cr
& = {{\mathop{\rm limn}\nolimits} ^2}{{\left( {1 + {1 \over n} – {3 \over {{n^4}}} – {2 \over {{n^5}}}} \right)} \over {\left( {4 + {6 \over n} + {9 \over {{n^3}}}} \right)}} = + \infty \cr} \)
c.
\(\eqalign{
& \lim {{\sqrt {2{n^4} + 3n – 2} } \over {2{n^2} – n + 3}} = \lim {{{n^2}\sqrt {2 + {3 \over {n^3}} – {2 \over {{n^4}}}} } \over {{n^2}\left ({2 – {1 \over n} + {3 \over{ {n^2}}}}\right )}} \cr
& = \lim {{\sqrt {2 + {n \over 3} – {2 \over {{n^2}}}} } \over {2 – {1 \over n} + {3 \over {{n^2}}}}} = {{\sqrt 2 } \over 2} \cr} \)
d. Chia cả tử và mẫu cho 5n ta được :
\(\eqalign{
& \lim {{{3^n} – {{2.5}^n}} \over {7 + {{3.5}^n}}} = \lim {{{{\left( {{3 \over 5}} \right)}^n} – 2} \over {7.{{\left( {{1 \over 5}} \right)}^n} + 3}} = – {2 \over 3} \cr
& \text{vì}\,\,\lim {\left( {{3 \over 5}} \right)^n} = \lim {\left( {{1 \over 5}} \right)^n} = 0 \cr} \)