\(sosanh\frac{n}{n+1}va\frac{3n+1}{6n+3}\)
K
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KN
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13 tháng 2 2017
\(\frac{n}{2n+1}\)=\(\frac{3.n}{3.\left(2n+1\right)}\)=\(\frac{3n}{6n+3}\)
Vì 6n+3=6n+3;3n<3n+1 nên \(\frac{n}{2n+1}\)<\(\frac{3n+1}{6n+3}\)
HQ
Hà Quang Minh
Giáo viên
22 tháng 9 2023
a) \(\lim \frac{{2{n^2} + 6n + 1}}{{8{n^2} + 5}} = \lim \frac{{{n^2}\left( {2 + \frac{6}{n} + \frac{1}{{{n^2}}}} \right)}}{{{n^2}\left( {8 + \frac{5}{{{n^2}}}} \right)}} = \lim \frac{{2 + \frac{6}{n} + \frac{1}{n}}}{{8 + \frac{5}{n}}} = \frac{2}{8} = \frac{1}{4}\)
b) \(\lim \frac{{4{n^2} - 3n + 1}}{{ - 3{n^3} + 6{n^2} - 2}} = \lim \frac{{{n^3}\left( {\frac{4}{n} - \frac{3}{{{n^2}}} + \frac{1}{{{n^3}}}} \right)}}{{{n^3}\left( { - 3 + \frac{6}{n} - \frac{2}{{{n^3}}}} \right)}} = \lim \frac{{\frac{4}{n} - \frac{3}{{{n^2}}} + \frac{1}{{{n^3}}}}}{{ - 3 + \frac{6}{n} - \frac{2}{{{n^3}}}}} = \frac{{0 - 0 + 0}}{{ - 3 + 0 - 0}} = 0\).
c) \(\lim \frac{{\sqrt {4{n^2} - n + 3} }}{{8n - 5}} = \lim \frac{{n\sqrt {4 - \frac{1}{n} + \frac{3}{{{n^2}}}} }}{{n\left( {8 - \frac{5}{n}} \right)}} = \frac{{\sqrt {4 - 0 + 0} }}{{8 - 0}} = \frac{2}{8} = \frac{1}{4}\).
d) \(\lim \left( {4 - \frac{{{2^{{\rm{n}} + 1}}}}{{{3^{\rm{n}}}}}} \right) = \lim \left( {4 - 2 \cdot {{\left( {\frac{2}{3}} \right)}^{\rm{n}}}} \right) = 4 - 2.0 = 4\).
e) \(\lim \frac{{{{4.5}^{\rm{n}}} + {2^{{\rm{n}} + 2}}}}{{{{6.5}^{\rm{n}}}}} = \lim \frac{{{{4.5}^{\rm{n}}} + {2^2}{{.2}^{\rm{n}}}}}{{{{6.5}^{\rm{n}}}}} = \lim \frac{{{5^n}.\left[ {4 + 4.{{\left( {\frac{2}{5}} \right)}^{\rm{n}}}} \right]}}{{{{6.5}^n}}} = \lim \frac{{4 + 4.{{\left( {\frac{2}{5}} \right)}^{\rm{n}}}}}{6} = \frac{{4 + 4.0}}{6} = \frac{2}{3}\).
g) \(\lim \frac{{2 + \frac{4}{{{n^3}}}}}{{{6^{\rm{n}}}}} = \lim \left( {2 + \frac{4}{{{{\rm{n}}^3}}}} \right).\lim {\left( {\frac{1}{6}} \right)^{\rm{n}}} = \left( {2 + 0} \right).0 = 0\).
HA
28 tháng 1 2016
Đặt A=1/2.5+1/5.8+...+1/(3n-1).(3n+2)
=>3A=3/2.5+3/5.8+...+3/(3n-1).(3n+2)
=>3A=1/2-1/5+1/5-1/8+...+1/3n-1-1/3n+2
=>3A=1/2-1/3n+2
=>3A=(3n+2-2)/[2.(3n+2)]
=>3A=3n/6n+4
=>A=3n/6n+4/3
=>A=n/6n+4
14 tháng 4 2016
Đặt \(A=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+......+\frac{1}{\left(3n-1\right)\left(3n+2\right)}\)
\(=>3A=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+....+\frac{3}{\left(3n-1\right)\left(3n+2\right)}\)
=> \(3A=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+....+\frac{1}{3n-1}-\frac{1}{3n+2}\)
=>\(3A=\frac{1}{2}-\frac{1}{3n+2}\)
=> \(3A=\frac{\left(3n+2\right):2}{3n+2}-\frac{1}{3n+2}\)
=> \(3A=\frac{1,5.n}{3n+2}\)
=>\(A=\frac{1,5.n}{3n+2}.\frac{1}{3}=>A=\frac{1,5.n}{\left(3n+2\right).3}=\frac{1,5.n}{9n+6}\)
\(Hay\) \(A=\frac{1,5n:1,5}{\left(9n+6\right):1,5}=\frac{n}{9n:1,5+6:1,5}=\frac{n}{6n + 4} \left(đpcm\right)\)