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GH
1
NV
Nguyễn Việt Lâm
Giáo viên
31 tháng 1 2021
Số 6 cuối cùng nằm trong căn bậc 3 hay ngoài căn bậc 3 bạn nhỉ?
NV
Nguyễn Việt Lâm
Giáo viên
12 tháng 1 2019
\(lim\dfrac{\left(n+2\right)^{50}\left(n-3\right)^{80}}{\left(2n-1\right)^{40}\left(3n-2\right)^{45}}=lim\dfrac{\left(1+\dfrac{2}{n^{50}}\right)\left(1-\dfrac{3}{n^{35}}\right)\left(n-3\right)^{45}}{\left(2-\dfrac{1}{n^{50}}\right)\left(3-\dfrac{2}{n^{45}}\right)}=+\infty\)
\(lim\dfrac{4^n}{2.3^n+4^n}=lim\dfrac{1}{2.\left(\dfrac{3}{4}\right)^n+1}=\dfrac{1}{0+1}=1\)
\(lim\dfrac{3^n-2.5^n}{7+3.5^n}=lim\dfrac{\left(\dfrac{3}{5}\right)^n-2}{\dfrac{7}{5^n}+3}=\dfrac{0-2}{0+3}=\dfrac{-2}{3}\)
\(lim\dfrac{4^n-5^n}{2^{2n}+3.5^{2n}}=lim\dfrac{\left(\dfrac{4}{25}\right)^n-\left(\dfrac{1}{5}\right)^n}{\left(\dfrac{2}{5}\right)^{2n}+3}=\dfrac{0-0}{0+3}=0\)
\(lim\dfrac{\left(-3\right)^n+5^n}{2.\left(-4\right)^n+5^n}=lim\dfrac{\left(\dfrac{-3}{5}\right)^n+1}{2.\left(-\dfrac{4}{5}\right)^n+1}=\dfrac{0+1}{0+1}=1\)
AH
Akai Haruma
Giáo viên
12 tháng 1 2019
1.
Nhớ rằng \(\lim _{x\to \infty}\frac{1}{x}=0\) và \(\lim _{x\to a}\frac{f(x)}{g(x)}=\frac{\lim_{x\to a}f(x)}{\lim_{x\to a}g(x)}\) với \(g(x)\neq 0; \lim_{x\to a}g(x)\neq 0\)
Do đó:
\(\lim_{n\to \infty}\frac{(n+2)^{50}.(n-3)^{80}}{(2n-1)^{40}.(3n-2)^{45}}=\lim_{n\to \infty}\frac{n^{130}(\frac{n+2}{n})^{50}.(\frac{n-3}{n})^{80}}{n^{85}(\frac{2n-1}{n})^{40}.(\frac{3n-2}{n})^{45}}\)
\(=\lim_{n\to \infty}\frac{n^{45}(1+\frac{2}{n})^{50}(1-\frac{3}{n})^{80}}{(2-\frac{1}{n})^{40}.(3-\frac{2}{n})^{45}}\)
\(=\frac{\lim_{n\to \infty}[n^{45}(1+\frac{2}{n})^{50}(1-\frac{3}{n})^{80}]}{\lim_{n\to \infty}[(2-\frac{1}{n})^{40}.(3-\frac{2}{n})^{45}]}\)
\(=\frac{\lim_{n\to \infty}n^{45}.1^{50}.1^{80}}{2^{40}.3^{45}}=\frac{\infty}{2^{40}.3^{45}}=\infty\)
10 tháng 3 2020
Đặt \(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+....+\frac{1}{n\left(n+1\right)}=A\)
\(\Leftrightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{n}-\frac{1}{n+1}\)
\(\Leftrightarrow A=\frac{n+1}{n+1}-\frac{1}{n+1}=\frac{n}{n+1}\)
NV
Nguyễn Việt Lâm
Giáo viên
5 tháng 1 2021
\(a=lim\dfrac{\left(\dfrac{2}{6}\right)^n+1-\dfrac{1}{4}\left(\dfrac{4}{6}\right)^n}{\left(\dfrac{3}{6}\right)^n+6}=\dfrac{1}{6}\)
\(b=\lim\dfrac{\left(n+1\right)^2}{3n^2+4}=\lim\dfrac{n^2+2n+1}{3n^2+4}=\lim\dfrac{1+\dfrac{2}{n}+\dfrac{1}{n^2}}{3+\dfrac{4}{n^2}}=\dfrac{1}{3}\)
\(c=\lim\dfrac{n\left(n+1\right)}{2\left(n^2-3\right)}=\lim\dfrac{n^2+n}{2n^2-6}=\lim\dfrac{1+\dfrac{1}{n}}{2-\dfrac{6}{n^2}}=\dfrac{1}{2}\)
\(d=\lim\left[1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n}-\dfrac{1}{n+1}\right]=\lim\left[1-\dfrac{1}{n+1}\right]=1\)
\(e=\lim\dfrac{1}{2}\left[1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right]\)
\(=\lim\dfrac{1}{2}\left[1-\dfrac{1}{2n+1}\right]=\dfrac{1}{2}\)
NV
Nguyễn Việt Lâm
Giáo viên
19 tháng 2 2020
a/ \(=lim\frac{1}{\sqrt{n+1}+\sqrt{n}}=\frac{1}{\infty}=0\)
b/ \(=lim\frac{6n+1}{\sqrt{n^2+5n+1}+\sqrt{n^2-n}}=\frac{6+\frac{1}{n}}{\sqrt{1+\frac{5}{n}+\frac{1}{n^2}}+\sqrt{1-\frac{1}{n}}}=\frac{6}{1+1}=3\)
c/ \(=lim\frac{6n-9}{\sqrt{3n^2+2n-1}+\sqrt{3n^2-4n+8}}=lim\frac{6-\frac{9}{n}}{\sqrt{3+\frac{2}{n}-\frac{1}{n^2}}+\sqrt{3-\frac{4}{n}+\frac{8}{n^2}}}=\frac{6}{\sqrt{3}+\sqrt{3}}=\sqrt{3}\)
d/ \(=lim\frac{\left(\frac{2}{6}\right)^n+1-4\left(\frac{4}{6}\right)^n}{\left(\frac{3}{6}\right)^n+6}=\frac{1}{6}\)
NV
Nguyễn Việt Lâm
Giáo viên
19 tháng 2 2020
e/ \(=lim\frac{\left(\frac{3}{5}\right)^n-\left(\frac{4}{5}\right)^n+1}{\left(\frac{3}{5}\right)^n+\left(\frac{4}{5}\right)^n-1}=\frac{1}{-1}=-1\)
f/ Ta có công thức:
\(1+3+...+\left(2n+1\right)^2=\left(n+1\right)^2\)
\(\Rightarrow lim\frac{1+3+...+2n+1}{3n^2+4}=lim\frac{\left(n+1\right)^2}{3n^2+4}=lim\frac{\left(1+\frac{1}{n}\right)^2}{3+\frac{4}{n^2}}=\frac{1}{3}\)
g/ \(=lim\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n}-\frac{1}{n+1}\right)=lim\left(1-\frac{1}{n+1}\right)=1-0=1\)
h/ Ta có: \(1^2+2^2+...+n^2=\frac{n\left(n+1\right)\left(2n+1\right)}{6}\)
\(\Rightarrow lim\frac{n\left(n+1\right)\left(2n+1\right)}{6n\left(n+1\right)\left(n+2\right)}=lim\frac{2n+1}{6n+12}=lim\frac{2+\frac{1}{n}}{6+\frac{12}{n}}=\frac{2}{6}=\frac{1}{3}\)
18 tháng 1 2021
b) \(lim\dfrac{4^n}{2.3^n+4^n}=lim\dfrac{1}{2.\left(\dfrac{3}{4}\right)^n+1}=\dfrac{1}{1}=1\).
NV
Nguyễn Việt Lâm
Giáo viên
18 tháng 1 2021
\(a=\lim\dfrac{17+\dfrac{3}{n}+\dfrac{4}{n^3}}{2+\dfrac{1}{n^2}}=\dfrac{17}{2}\)
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) \(\lim \frac{{2 + \frac{4}{{{n^3}}}}}{{{6^{\rm{n}}}}} = \lim \left( {2 + \frac{4}{{{{\rm{n}}^3}}}} \right).\lim {\left( {\frac{1}{6}} \right)^{\rm{n}}} = \left( {2 + 0} \right).0 = 0\).
NV
Nguyễn Việt Lâm
Giáo viên
13 tháng 2 2022
\(\lim\dfrac{\left(-3\right)^n-4.5^{n+1}}{2.4^n+3.5^n}=\lim\dfrac{\left(-3\right)^n+20.5^n}{2.4^n+3.5^n}=\lim\dfrac{\left(-\dfrac{3}{5}\right)^n+20}{2\left(\dfrac{4}{5}\right)^n+3}=\dfrac{0+20}{0+3}=\dfrac{20}{3}\)
\(\lim\dfrac{2^n-3^n+4.5^{n+2}}{2^{n+1}+3^{n+2}+5^{n+1}}=\lim\dfrac{2^n-3^n+100.5^n}{2.2^n+9.3^n+5.5^n}=\lim\dfrac{\left(\dfrac{2}{5}\right)^n-\left(\dfrac{3}{5}\right)^n+100}{2\left(\dfrac{2}{5}\right)^n+9\left(\dfrac{3}{5}\right)^n+5}=\dfrac{100}{5}=20\)
26 tháng 8 2023
a) \(\lim\limits\dfrac{2n^2+3n}{n^2+1}=\lim\limits\dfrac{n^2\left(2+\dfrac{3n}{n^2}\right)}{n^2\left(1+\dfrac{1}{n^2}\right)}=\lim\limits\dfrac{2+\dfrac{3}{n}}{1+\dfrac{1}{n^2}}=2\).
b) \(\lim\limits\dfrac{\sqrt{4n^2+3}}{n}\\ =\lim\limits\dfrac{\sqrt{n^2\left(4+\dfrac{3}{n^2}\right)}}{n}\\ =\lim\limits\dfrac{\sqrt[n]{4+\dfrac{3}{n^2}}}{n}\\ =\lim\limits\sqrt{4+\dfrac{3}{n^2}}\\ =2.\)
lim\(\frac{2^n+4^n+5^n}{2.3^n+4^n+3.5^n}\)
=lim\(\frac{\left(\frac{2}{5}\right)^n+\left(\frac{4}{5}\right)^n+1}{2.\left(\frac{3}{5}\right)^n+\left(\frac{4}{5}\right)^n-3}=-\frac{1}{3}\)