Bài 6. Tìm \(\lim{\rm{ }}{u_n}\) với
a. \({u_n} = {{{n^2} - 3n + 5} \over {2{n^2} - 1}}\)
b. \({u_n} = {{ - 2{n^2} + n + 2} \over {3{n^4} + 5}}\)
c. \({u_n} = {{\sqrt {2{n^2} - n} } \over {1 - 3{n^2}}}\)
d. \({u_n} = {{{4^n}} \over {{{2.3}^n} + {4^n}}}\)
a. Ta có:
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\(\eqalign{
& \lim{u_n} = \lim {{{n^2}\left( {1 - {3 \over n} + {5 \over {{n^2}}}} \right)} \over {{n^2}\left( {2 - {1 \over {{n^2}}}} \right)}} = \lim {{1 - {3 \over n} + {5 \over {{n^2}}}} \over {2 - {1 \over {{n^2}}}}} \cr
& = {{\lim 1 - \lim {3 \over n} + \lim {5 \over {{n^2}}}} \over {\lim 2 - \lim {1 \over {{n^2}}}}} = {{1 - 0 + 0} \over {2 - 0}} = {1 \over 2} \cr} \)
b.
\(\lim {u_n} = \lim {{{n^4}\left( {{{ - 2} \over {{n^2}}} + {1 \over {{n^3}}} + {{ 2} \over {{n^4}}}} \right)} \over {{n^4}\left( {3 + {5 \over {{n^4}}}} \right)}} = \lim {{{{ - 2} \over {{n^2}}} + {1 \over {{n^3}}} + {{ 2} \over {{n^4}}}} \over {3 + {5 \over {{n^4}}}}} = {0 \over 3} = 0\)
c.
\({{\mathop{\rm limu}\nolimits} _n} = \lim {{{n^2}\sqrt {{2 \over {{n^2}}} - {1 \over {{n^3}}}} } \over {{n^2}\left( {{1 \over {{n^2}}} - 3} \right)}} = \lim {{\sqrt {{2 \over {{n^2}}} - {1 \over {{n^3}}}} } \over {{1 \over {{n^2}}} - 3}} = {0 \over { - 3}} = 0\)
d. Chia cả tử và mẫu \(u_n\) cho \(4^n\) ta được :
\(\lim {u_n} = \lim {1 \over {2.{{\left( {{3 \over 4}} \right)}^n} + 1}} = 1\,\text{ vì }\,\lim {\left( {{3 \over 4}} \right)^n} = 0\)