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Tìm giới hạn của các dãy số (un) với
a. \({u_n} = {{2{n^3} – n – 3} \over {5n – 1}}\)
b. \({u_n} = {{\sqrt {{n^4} – 2n + 3} } \over { – 2{n^2} + 3}}\)
c. \({u_n} = – 2{n^2} + 3n – 7\)
d. \({u_n} = \root 3 \of {{n^9} + 8{n^2} – 7} \)
a. Ta có:
\(\eqalign{
& \lim {{2{n^3} – n – 3} \over {5n – 1}} = \lim {{{n^3}\left( {2 – {1 \over {{n^2}}} – {3 \over {{n^3}}}} \right)} \over {{n^3}\left( {{5 \over {{n^2}}} – {1 \over {{n^3}}}} \right)}} \cr
& = \lim {{2 – {1 \over {{n^2}}} – {3 \over {{n^3}}}} \over {{5 \over {{n^2}}} – {1 \over {{n^3}}}}} = + \infty \cr
& \text{ vì }\,\lim \left( {2 – {1 \over {{n^2}}} – {3 \over {{n^3}}}} \right) = 2\,\text{ và }\,\lim \left( {{5 \over {{n^2}}} – {1 \over {{n^3}}}} \right) = 0;5n – 1 > 0 \cr} \)
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b.
\(\eqalign{
& \lim {{\sqrt {{n^4} – 2n + 3} } \over { – 2{n^2} + 3}} = \lim {{{n^2}\sqrt {1 – {2 \over {{n^3}}} + {3 \over {{n^4}}}} } \over {{n^2}\left( { – 2 + {3 \over {{n^2}}}} \right)}} \cr
& = \lim {{\sqrt {1 – {2 \over {{n^3}}} + {3 \over {{n^4}}}} } \over { – 2 + {3 \over {{n^2}}}}} = – {1 \over 2} \cr} \)
c.
\(\eqalign{
& \lim \left( { – 2{n^2} + 3n – 7} \right) = \lim {n^2}\left( { – 2 + {3 \over n} – {7 \over {{n^2}}}} \right) = – \infty \cr
& \text{vì }\,\lim {n^2} = + \infty \,\text{ và }\,\lim \left( { – 2 + {3 \over n} – {7 \over {{n^2}}}} \right) = – 2 < 0 \cr} \)
d.
\(\eqalign{
& \lim \root 3 \of {{n^9} + 8{n^2} – 7} = \lim {n^3}.\root 3 \of {1 + {8 \over {{n^7}}} – {7 \over {{n^9}}}} = + \infty \cr
& \text{ vì }\,\lim {n^3} = + \infty \,\text{ và }\,\lim \root 3 \of {1 + {8 \over {{n^7}}} – {7 \over {{n^9}}}} = 1 > 0 \cr} \)