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Bài 10. Tính các giới hạn sau
a) \(\lim {{(n + 1){{(3 – 2n)}^2}} \over {{n^3} + 1}}\)
b) \(\lim ({1 \over {{n^2} + 1}} + {2 \over {{n^2} + 1}} + {3 \over {{n^2} + 1}} + … + {{n – 1} \over {{n^2} + 1}})\)
c) \(\lim {{\sqrt {4n + 1} + n} \over {2n + 1}}\)
d) \(\lim \sqrt n (\sqrt {n – 1} – \sqrt n )\)
a)
\(\eqalign{
& \lim {{(n + 1){{(3 – 2n)}^2}} \over {{n^3} + 1}} = \lim {{(1 + {1 \over n}){{({3 \over n} – 2)}^2}} \over {1 + {1 \over {{n^3}}}}} \cr
& = {{(1 + 0){{(0 – 2)}^2}} \over {1 + 0}} = 4 \cr} \)
b)
\(\eqalign{
& {1 \over {{n^2} + 1}} + {2 \over {{n^2} + 1}} + {3 \over {{n^2} + 1}} + … + {{n – 1} \over {{n^2} + 1}} \cr
& = {{1 + 2 + … + n – 1} \over {{n^2} + 1}} \cr
& = {{{{n(n – 1)} \over 2}} \over {{n^2} + 1}} = {{{n^2} -n} \over {2({n^2} + 1)}} \cr
& \Rightarrow \lim ({1 \over {{n^2} + 1}} + {2 \over {{n^2} + 1}} + {3 \over {{n^2} + 1}} + … + {{n – 1} \over {{n^2} + 1}}) \cr
& = lim{{{n^2} -n} \over {2({n^2} + 1)}} \cr
& = \lim {{{n^2}(1 – {1 \over n} )} \over {2{n^2}(1 + {1 \over {{n^2}}})}} \cr
& = \lim {{1 – {1 \over n} } \over {2(1 + {1 \over {{n^2}}})}} = {1 \over 2} \cr} \)
c)
\(\eqalign{
& \lim {{\sqrt {4n^2 + 1} + n} \over {2n + 1}} \cr
& = \lim {{n.\sqrt {4 + {1 \over {{n^2}}}} + n} \over {2n + 1}} \cr
& = \lim {{n.(\sqrt {4 + {1 \over {{n^2}}}} + 1)} \over {n(2 + {1 \over n})}} \cr
& = \lim {{\sqrt {4 + {1 \over {{n^2}}}} + 1} \over {2 + {1 \over n}}} \cr
& = {{2 + 1} \over 2} = {3 \over 2} \cr} \)
d)
\(\eqalign{
& \lim \sqrt n (\sqrt {n – 1} – \sqrt n ) \cr
& = \lim {{\sqrt n (\sqrt {n – 1} – \sqrt n )(\sqrt {n – 1} + \sqrt n )} \over {\sqrt {n – 1} + \sqrt n }} \cr
& = \lim {{\sqrt n \left[ {(n – 1) – n} \right]} \over {\sqrt {n – 1} + \sqrt n }} \cr
& = \lim {{ – \sqrt n } \over {\sqrt n \left[ {\sqrt {1 – {1 \over n}} + 1} \right]}} \cr
& = \lim {{ – 1} \over {\sqrt {1 – {1 \over n}} + 1}} = – {1 \over 2} \cr} \)