Giải bài 10 Ôn tập cuối năm

a. \(\lim {{(n + 1){{(3 - 2n)}^2}} \over {{n^3} + 1}} \)

\(= \lim {{(1 + {1 \over n}){{({3 \over n} - 2)}^2}} \over {1 + {1 \over {{n^3}}}}} \)

\(= {{(1 + 0){{(0 - 2)}^2}} \over {1 + 0}} = 4\)

b. Ta có:

\({1 \over {{n^2} + 1}} + {2 \over {{n^2} + 1}} + {3 \over {{n^2} + 1}} + ... + {{n - 1} \over {{n^2} + 1}} \)

\(= {{1 + 2 + ... + n - 1} \over {{n^2} + 1}} \)

\(= {{{{n(n - 1)} \over 2}} \over {{n^2} + 1}} \)

\(= {{{n^2} -n} \over {2({n^2} + 1)}} \)

\(\Rightarrow \lim ({1 \over {{n^2} + 1}} + {2 \over {{n^2} + 1}} + {3 \over {{n^2} + 1}} + ... + {{n - 1} \over {{n^2} + 1}}) \)

\(= lim{{{n^2} -n} \over {2({n^2} + 1)}} \)

\(= \lim {{{n^2}(1 - {1 \over n} )} \over {2{n^2}(1 + {1 \over {{n^2}}})}} \)

\(= \lim {{1 - {1 \over n} } \over {2(1 + {1 \over {{n^2}}})}} = {1 \over 2} \)

c. \(\lim {{\sqrt {4n^2 + 1} + n} \over {2n + 1}} \)

\(= \lim {{n.\sqrt {4 + {1 \over {{n^2}}}} + n} \over {2n + 1}} \)

\(= \lim {{n.\left ( \sqrt {4 + {1 \over {{n^2}}}} + 1 \right )} \over {n(2 + {1 \over n})}} \)

\(= \lim {{\sqrt {4 + {1 \over {{n^2}}}} + 1} \over {2 + {1 \over n}}} \)

\(= {{2 + 1} \over 2} = {3 \over 2}\)

d. \(\lim \sqrt n (\sqrt {n - 1} - \sqrt n ) \)

\(= \lim {{\sqrt n (\sqrt {n - 1} - \sqrt n )(\sqrt {n - 1} + \sqrt n )} \over {\sqrt {n - 1} + \sqrt n }} \)

\(= \lim {{\sqrt n \left[ {(n - 1) - n} \right]} \over {\sqrt {n - 1} + \sqrt n }} \)

\(= \lim {{ - \sqrt n } \over {\sqrt n \left( {\sqrt {1 - {1 \over n}} + 1} \right)}} \)

\(= \lim {{ - 1} \over {\sqrt {1 - {1 \over n}} + 1}} = - {1 \over 2}\)