lim 1 2 + 2 2 + . . . + n 2 3 n bằng
A. 1
B. + ∞
C. 3 3 3
D. 1 3 3
K
Khách
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HT
1 tháng 2 2021
a/ \(=\lim\limits\dfrac{\sqrt{\dfrac{n}{n}+\dfrac{1}{n}}}{\dfrac{1}{\sqrt{n}}+\sqrt{\dfrac{n}{n}}}=1\)
b/ \(1+2+...+n=\dfrac{n\left(n+1\right)}{2}\)
\(\Rightarrow\lim\limits\dfrac{n\left(n+1\right)}{2n^2+4}=\lim\limits\dfrac{\dfrac{n^2}{n^2}+\dfrac{n}{n^2}}{\dfrac{2n^2}{n^2}+\dfrac{4}{n^2}}=\dfrac{1}{2}\)
c/ \(=\lim\limits\dfrac{n^2+n+1-n^2}{\sqrt{n^2+n+1}+n}=\lim\limits\dfrac{n+1}{\sqrt{n^2+n+1}+n}=\lim\limits\dfrac{\dfrac{n}{n}+\dfrac{1}{n}}{\sqrt{\dfrac{n^2}{n^2}+\dfrac{n}{n^2}+\dfrac{1}{n^2}}+\dfrac{n}{n}}=\dfrac{1}{1+1}=\dfrac{1}{2}\)
d/ \(=\lim\limits\left[\sqrt{n}\left(\sqrt{3-\dfrac{1}{\sqrt{n}}}-\sqrt{2-\dfrac{1}{\sqrt{n}}}\right)\right]=\lim\limits\left[\sqrt{n}\left(\sqrt{3}-\sqrt{2}\right)\right]=+\infty\)
e/ \(=\lim\limits\dfrac{n^3+2n^2-n-n^3}{\left(\sqrt[3]{n^3+2n^2}\right)^2+n.\sqrt[3]{n^3+2n^2}+n^2}=\lim\limits\dfrac{2n^2-n}{\left(n^3+2n^2\right)^{\dfrac{2}{3}}+n.\left(n^3+2n^2\right)^{\dfrac{1}{3}}+n^2}\)
\(=\dfrac{2}{1+1+1}=\dfrac{2}{3}\)
g/ \(=\lim\limits\dfrac{2^n+9.3^n}{4.3^n+8.2^n}=\lim\limits\dfrac{\left(\dfrac{2}{3}\right)^n+9.\left(\dfrac{3}{3}\right)^n}{4.\left(\dfrac{3}{3}\right)^n+8.\left(\dfrac{2}{3}\right)^n}=\dfrac{9}{4}\)
HQ
Hà Quang Minh
Giáo viên
22 tháng 9 2023
\(\lim \frac{{n + 3}}{{{n^2}}} = \lim \frac{{{n^2}\left( {\frac{1}{n} + \frac{3}{{{n^2}}}} \right)}}{{{n^2}}} = \lim \left( {\frac{1}{n} + \frac{3}{{{n^2}}}} \right) = 0\)
Chọn B.
NV
Nguyễn Việt Lâm
Giáo viên
20 tháng 1 2021
\(a=\lim\sqrt{n^3}\sqrt{\dfrac{1}{n^3}+\dfrac{2}{n^2}-1}=\infty.\left(-1\right)=-\infty\)
\(b=\lim\left(\sqrt{n^2+2n+3}-n+n-\sqrt[3]{n^2+n^3}\right)\)
\(=\lim\dfrac{2n+3}{\sqrt{n^2+2n+3}+n}+\lim\dfrac{-n^2}{n^2+n\sqrt[3]{n^2+n^3}+\sqrt[3]{\left(n^2+n^3\right)^2}}\)
\(=\lim\dfrac{2+\dfrac{3}{n}}{\sqrt{1+\dfrac{2}{n}+\dfrac{3}{n^2}}+1}+\lim\dfrac{-1}{1+\sqrt[3]{\dfrac{1}{n}+1}+\sqrt[3]{\left(\dfrac{1}{n}+1\right)^2}}=\dfrac{2}{2}-\dfrac{1}{3}=\dfrac{2}{3}\)
\(c=\lim\dfrac{\left(\dfrac{2}{\sqrt{n}}+\dfrac{1}{n}\right)\left(\dfrac{1}{\sqrt{n}}+\dfrac{3}{n}\right)}{\left(1+\dfrac{1}{n}\right)\left(1+\dfrac{2}{n}\right)}=\dfrac{0.0}{1.1}=0\)
NV
Nguyễn Việt Lâm
Giáo viên
20 tháng 1 2021
\(d=\lim\dfrac{4-3\left(\dfrac{2}{4}\right)^n}{9.\left(\dfrac{3}{4}\right)^n+\left(\dfrac{2}{4}\right)^n}=\dfrac{4}{0}=+\infty\)
\(e=\lim\dfrac{7-25\left(\dfrac{5}{7}\right)^n+3.\left(\dfrac{1}{7}\right)^n}{12.\left(\dfrac{6}{7}\right)^n-\left(\dfrac{3}{7}\right)^n+3\left(\dfrac{1}{7}\right)^n}=\dfrac{7}{0}=+\infty\)
\(f=\lim\dfrac{n^4-4n^6}{n\left(\sqrt{n^4+1}+\sqrt{4n^6+1}\right)}=\lim\dfrac{\dfrac{1}{n^2}-6}{\sqrt{\dfrac{1}{n^6}+\dfrac{1}{n^{10}}}+\sqrt{\dfrac{4}{n^4}+\dfrac{1}{n^{10}}}}=\dfrac{-6}{0}=-\infty\)
4 tháng 11 2023
a: \(\lim\limits\dfrac{5n+1}{2n}=\lim\limits\dfrac{\dfrac{5n}{n}+\dfrac{1}{n}}{\dfrac{2n}{n}}=\lim\limits\dfrac{5+\dfrac{1}{n}}{2}=\dfrac{5+0}{2}=\dfrac{5}{2}\)
b: \(\lim\limits\dfrac{6n^2+8n+1}{5n^2+3}\)
\(=\lim\limits\dfrac{\dfrac{6n^2}{n^2}+\dfrac{8n}{n^2}+\dfrac{1}{n^2}}{\dfrac{5n^2}{n^2}+\dfrac{3}{n^2}}\)
\(=\lim\limits\dfrac{6+\dfrac{8}{n}+\dfrac{1}{n^2}}{5+\dfrac{3}{n^2}}\)
\(=\dfrac{6+0+0}{5+0}=\dfrac{6}{5}\)
c: \(\lim\limits\dfrac{3^n+2^n}{4\cdot3^n}\)
\(=\lim\limits\dfrac{\dfrac{3^n}{3^n}+\left(\dfrac{2}{3}\right)^n}{4\cdot\left(\dfrac{3^n}{3^n}\right)}\)
\(=\lim\limits\dfrac{1+\left(\dfrac{2}{3}\right)^n}{4}=\dfrac{1+0}{4}=\dfrac{1}{4}\)
d: \(\lim\limits\dfrac{\sqrt{n^2+5n+3}}{6n+2}\)
\(=\lim\limits\dfrac{\sqrt{\dfrac{n^2}{n^2}+\dfrac{5n}{n^2}+\dfrac{3}{n^2}}}{\dfrac{6n}{n}+\dfrac{2}{n}}\)
\(=\lim\limits\dfrac{\sqrt{1+\dfrac{5}{n}+\dfrac{3}{n^2}}}{6+\dfrac{2}{n}}\)
\(=\dfrac{\sqrt{1+0+0}}{6}=\dfrac{1}{6}\)
AD
4 tháng 11 2023
\(a,lim\dfrac{5n+1}{2n}=lim\dfrac{\dfrac{5n}{n}+\dfrac{1}{n}}{\dfrac{2n}{n}}=lim\dfrac{5+\dfrac{1}{n}}{2}=\dfrac{5}{2}\\ b,lim\dfrac{6n^2+8n+1}{5n^2+3}=lim\dfrac{\dfrac{6n^2}{n^2}+\dfrac{8n}{n^2}+\dfrac{1}{n^2}}{\dfrac{5n^2}{n^2}+\dfrac{3}{n^2}}=lim\dfrac{6+\dfrac{8}{n}+\dfrac{1}{n^2}}{5+\dfrac{3}{n^2}}=\dfrac{6}{5}\)
\(c,lim\dfrac{3^n+2^n}{4.3^n}=\dfrac{\dfrac{3^n}{3^n}+\dfrac{2^n}{3^n}}{\dfrac{4.3^n}{3^n}}=\dfrac{1+\left(\dfrac{2}{3}\right)^n}{4}=\dfrac{1}{4}\)
\(d,lim\dfrac{\sqrt{n^2+5n+3}}{6n+2}=lim\dfrac{\sqrt{\dfrac{n^2+5n+3}{n^2}}}{\dfrac{6n}{n}+\dfrac{2}{n}}=lim\dfrac{\sqrt{1+\dfrac{5}{n}+\dfrac{3}{n^2}}}{6+\dfrac{2}{n}}=\dfrac{1}{6}\)
HQ
Hà Quang Minh
Giáo viên
22 tháng 9 2023
a) \(\lim \frac{{5n + 1}}{{2n}} = \lim \frac{{5 + \frac{1}{n}}}{2} = \frac{{5 + 0}}{2} = \frac{5}{2}\)
b) \(\lim \frac{{6{n^2} + 8n + 1}}{{5{n^2} + 3}} = \lim \frac{{6 + \frac{8}{n} + \frac{1}{{{n^2}}}}}{{5 + \frac{3}{{{n^2}}}}} = \frac{{6 + 0 + 0}}{{5 + 0}} = \frac{6}{5}\)
c) \(\lim \frac{{\sqrt {{n^2} + 5n + 3} }}{{6n + 2}} = \lim \frac{{\sqrt {1 + \frac{5}{n} + \frac{3}{{{n^2}}}} }}{{6 + \frac{2}{n}}} = \frac{{\sqrt {1 + 0 + 0} }}{{6 + 0}} = \frac{1}{6}\)
d) \(\lim \left( {2 - \frac{1}{{{3^n}}}} \right) = \lim 2 - \lim {\left( {\frac{1}{3}} \right)^n} = 2 - 0 = 0\)
e) \(\lim \frac{{{3^n} + {2^n}}}{{{{4.3}^n}}} = \lim \frac{{1 + {{\left( {\frac{2}{3}} \right)}^n}}}{4} = \frac{{1 + 0}}{4} = \frac{1}{4}\)
g) \(\lim \frac{{2 + \frac{1}{n}}}{{{3^n}}}\)
Ta có \(\lim \left( {2 + \frac{1}{n}} \right) = \lim 2 + \lim \frac{1}{n} = 2 + 0 = 2 > 0;\lim {3^n} = + \infty \Rightarrow \lim \frac{{2 + \frac{1}{n}}}{{{3^n}}} = 0\)
NV
Nguyễn Việt Lâm
Giáo viên
13 tháng 1 2021
\(a=\lim\dfrac{\dfrac{1}{n}+\dfrac{1}{n^2}}{1+\dfrac{2}{n}}=\dfrac{0}{1}=0\)
\(b=\lim n^3\left(-2+\dfrac{1}{n}+\dfrac{2}{n^3}\right)=+\infty.\left(-2\right)=-\infty\)
\(c=\lim\dfrac{\sqrt{9-\dfrac{1}{n}-\dfrac{1}{n^2}}}{4-\dfrac{2}{n}}=\dfrac{\sqrt{9}}{4}=\dfrac{3}{4}\)
\(d=\lim\dfrac{\left(\dfrac{3}{4}\right)^n+5}{1+\left(\dfrac{2}{4}\right)^n}=\dfrac{5}{1}=5\)
NV
Nguyễn Việt Lâm
Giáo viên
6 tháng 2 2021
\(a=\lim n\left(\sqrt[3]{-1+\dfrac{2}{n}-\dfrac{5}{n^3}}\right)=+\infty.\left(-1\right)=-\infty\)
\(b=\lim\left(\sqrt{n+1}+\sqrt{n}\right)=+\infty\)
\(c=\lim n\left(\dfrac{1}{n^2+n}-1\right)=+\infty.\left(-1\right)=-\infty\)
\(d=\lim\left(\dfrac{2n^2-1-2n\left(n+1\right)}{n+1}\right)=\lim\left(\dfrac{-1-2n}{n+1}\right)=-2\)
\(e=\lim\dfrac{2n^2+n-3+\dfrac{1}{n}}{\dfrac{2}{n}-3}=\dfrac{+\infty}{-3}=-\infty\)
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) \(\lim \frac{{2 + \frac{4}{{{n^3}}}}}{{{6^{\rm{n}}}}} = \lim \left( {2 + \frac{4}{{{{\rm{n}}^3}}}} \right).\lim {\left( {\frac{1}{6}} \right)^{\rm{n}}} = \left( {2 + 0} \right).0 = 0\).
Đáp án D