Tính lim n → ∞ 5 . 3 n

QT

Quoc Tran Anh Le

Giáo viên

22 tháng 9 2023

Tính các giới hạn sau:a) $\lim \frac{{2{n^2} + 6n + 1}}{{8{n^2} + 5}}$ b) $\lim \frac{{4{n^2} - 3n + 1}}{{ - 3{n^3} + 5{n^2} - 2}}$; c) $\lim \frac{{\sqrt {4{n^2} - n + 3} }}{{8n - 5}}$;d) $\lim \left( {4 - \frac{{{2^{n + 1}}}}{{{3^n}}}} \right)$ e) $\lim \frac{{{{4.5}^n} + {2^{n + 2}}}}{{{{6.5}^n}}}$ g) \(\lim \frac{{2 +...

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#Toán lớp 11

1

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) $\lim \frac{{2 + \frac{4}{{{n^3}}}}}{{{6^{\rm{n}}}}} = \lim \left( {2 + \frac{4}{{{{\rm{n}}^3}}}} \right).\lim {\left( {\frac{1}{6}} \right)^{\rm{n}}} = \left( {2 + 0} \right).0 = 0$.

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HC

Hiếu Chuối

5 tháng 1 2021

Tìm giới hạn các dãy số...

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#Toán lớp 11

1

NV

Nguyễn Việt Lâm

Giáo viên

5 tháng 1 2021

$a=lim\dfrac{\left(\dfrac{2}{6}\right)^n+1-\dfrac{1}{4}\left(\dfrac{4}{6}\right)^n}{\left(\dfrac{3}{6}\right)^n+6}=\dfrac{1}{6}$

$b=\lim\dfrac{\left(n+1\right)^2}{3n^2+4}=\lim\dfrac{n^2+2n+1}{3n^2+4}=\lim\dfrac{1+\dfrac{2}{n}+\dfrac{1}{n^2}}{3+\dfrac{4}{n^2}}=\dfrac{1}{3}$

$c=\lim\dfrac{n\left(n+1\right)}{2\left(n^2-3\right)}=\lim\dfrac{n^2+n}{2n^2-6}=\lim\dfrac{1+\dfrac{1}{n}}{2-\dfrac{6}{n^2}}=\dfrac{1}{2}$

$d=\lim\left[1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n}-\dfrac{1}{n+1}\right]=\lim\left[1-\dfrac{1}{n+1}\right]=1$

$e=\lim\dfrac{1}{2}\left[1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right]$

$=\lim\dfrac{1}{2}\left[1-\dfrac{1}{2n+1}\right]=\dfrac{1}{2}$

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JE

Julian Edward

6 tháng 2 2021

Tìm các giới hạn sau:

a) $lim\left(4^n-3^n\right)$

b) $lim\left[\left(2^n+1\right)^2-4^n\right]$

c) $lim\left(\sqrt{2n^5-3n^2+11}-n^3\right)$

d) $lim\left(\sqrt{2n^2+1}-\sqrt{3n^2-1}\right)$

e) $lim\sqrt{n^2+3n\sqrt{n}+1}-n$

#Toán lớp 11

1

NV

Nguyễn Việt Lâm

Giáo viên

7 tháng 2 2021

$a=\lim4^n\left(1-\left(\dfrac{3}{4}\right)^n\right)=+\infty.1=+\infty$

$b=\lim\left(4^n+2.2^n+1-4^n\right)=\lim2^n\left(2+\dfrac{1}{2^n}\right)=+\infty.2=+\infty$

$c=limn^3\left(\sqrt{\dfrac{2}{n}-\dfrac{3}{n^4}+\dfrac{11}{n^6}}-1\right)=+\infty.\left(-1\right)=-\infty$

$d=\lim n\left(\sqrt{2+\dfrac{1}{n^2}}-\sqrt{3-\dfrac{1}{n^2}}\right)=+\infty\left(\sqrt{2}-\sqrt{3}\right)=-\infty$

$e=\lim\dfrac{3n\sqrt{n}+1}{\sqrt{n^2+3n\sqrt{n}+1}+n}=\lim\dfrac{3\sqrt{n}+\dfrac{1}{n}}{\sqrt{1+\dfrac{3}{\sqrt{n}}+\dfrac{1}{n^2}}+1}=\dfrac{+\infty}{2}=+\infty$

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NL

nguyễn linh chi

19 tháng 2 2020

tìm các giới...

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#Toán lớp 11

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/ $=lim\frac{\left(\frac{2}{6}\right)^n+1-4\left(\frac{4}{6}\right)^n}{\left(\frac{3}{6}\right)^n+6}=\frac{1}{6}$

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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: $1^2+2^2+...+n^2=\frac{n\left(n+1\right)\left(2n+1\right)}{6}$

$\Rightarrow lim\frac{n\left(n+1\right)\left(2n+1\right)}{6n\left(n+1\right)\left(n+2\right)}=lim\frac{2n+1}{6n+12}=lim\frac{2+\frac{1}{n}}{6+\frac{12}{n}}=\frac{2}{6}=\frac{1}{3}$

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T

títtt

15 tháng 10 2023

1) Tính giới hạn $\lim\limits_{n\rightarrow\infty}\left(\dfrac{3^n-4^{n+1}}{3^{n+2}+4^n}\right)$

2) Tính giới hạn $\lim\limits_{n\rightarrow\infty}\left(\dfrac{3^n-4.2^{n+1}-3}{3.2^n+4^n}\right)$

3) Tính giới hạn $\lim\limits_{n\rightarrow\infty}\left(\dfrac{2-5^{n-2}}{3^n+2.5^n}\right)$

#Toán lớp 11

1

NL

Nguyễn Lê Phước Thịnh

15 tháng 10 2023

3:

$\lim\limits_{n\rightarrow\infty}\dfrac{2-5^{n-2}}{3^n+2\cdot5^n}$

$=\lim\limits_{n\rightarrow\infty}\dfrac{\dfrac{2}{5^n}-\dfrac{5^{n-2}}{5^n}}{\dfrac{3^n}{5^n}+2\cdot\dfrac{5^n}{5^n}}$

$=\lim\limits_{n\rightarrow\infty}\dfrac{\dfrac{2}{5^n}-\dfrac{1}{25}}{\left(\dfrac{3}{5}\right)^n+2\cdot1}$

$=-\dfrac{1}{25}:2=-\dfrac{1}{50}$

1:

$=\lim\limits_{n\rightarrow\infty}\dfrac{3^n-4^n\cdot4}{3^n\cdot9+4^n}$

$=\lim\limits_{n\rightarrow\infty}\dfrac{\dfrac{3^n}{4^n}-4}{3^n\cdot\dfrac{9}{4^n}+1}$

$=-\dfrac{4}{1}=-4$

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NH

Nguyễn Hải Vân

1 tháng 2 2021

a, lim $\dfrac{\sqrt{n+1}}{1+\sqrt{n}}$

b, lim $\dfrac{1+2+...+n}{n^2+2}$

c, lim $(\sqrt{n^2+n+1}-n)$

d, lim $(\sqrt{3n-1}-\sqrt{2n-1})$

e, lim $(\sqrt[3]{n^3+2n^2}-n)$

g, lim $\dfrac{(2)^{n}+(3)^{n+2}}{4×(3)^{n}+(2)^{n+3}}$

#Toán lớp 11

1

HT

Hoàng Tử Hà

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/ $=\lim\limits\dfrac{2^n+9.3^n}{4.3^n+8.2^n}=\lim\limits\dfrac{\left(\dfrac{2}{3}\right)^n+9.\left(\dfrac{3}{3}\right)^n}{4.\left(\dfrac{3}{3}\right)^n+8.\left(\dfrac{2}{3}\right)^n}=\dfrac{9}{4}$

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NH

Nguyễn Hải Vân

1 tháng 2 2021

Mình cảm ơn nhiều nhé❤

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TN

Thư Nguyễn Huỳnh Anh

20 tháng 1 2021

...

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#Toán lớp 11

2

NV

Nguyễn Việt Lâm

Giáo viên

20 tháng 1 2021

$a=\lim\sqrt{n^3}\sqrt{\dfrac{1}{n^3}+\dfrac{2}{n^2}-1}=\infty.\left(-1\right)=-\infty$

$b=\lim\left(\sqrt{n^2+2n+3}-n+n-\sqrt[3]{n^2+n^3}\right)$

$=\lim\dfrac{2n+3}{\sqrt{n^2+2n+3}+n}+\lim\dfrac{-n^2}{n^2+n\sqrt[3]{n^2+n^3}+\sqrt[3]{\left(n^2+n^3\right)^2}}$

$=\lim\dfrac{2+\dfrac{3}{n}}{\sqrt{1+\dfrac{2}{n}+\dfrac{3}{n^2}}+1}+\lim\dfrac{-1}{1+\sqrt[3]{\dfrac{1}{n}+1}+\sqrt[3]{\left(\dfrac{1}{n}+1\right)^2}}=\dfrac{2}{2}-\dfrac{1}{3}=\dfrac{2}{3}$

$c=\lim\dfrac{\left(\dfrac{2}{\sqrt{n}}+\dfrac{1}{n}\right)\left(\dfrac{1}{\sqrt{n}}+\dfrac{3}{n}\right)}{\left(1+\dfrac{1}{n}\right)\left(1+\dfrac{2}{n}\right)}=\dfrac{0.0}{1.1}=0$

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NV

Nguyễn Việt Lâm

Giáo viên

20 tháng 1 2021

$d=\lim\dfrac{4-3\left(\dfrac{2}{4}\right)^n}{9.\left(\dfrac{3}{4}\right)^n+\left(\dfrac{2}{4}\right)^n}=\dfrac{4}{0}=+\infty$

$e=\lim\dfrac{7-25\left(\dfrac{5}{7}\right)^n+3.\left(\dfrac{1}{7}\right)^n}{12.\left(\dfrac{6}{7}\right)^n-\left(\dfrac{3}{7}\right)^n+3\left(\dfrac{1}{7}\right)^n}=\dfrac{7}{0}=+\infty$

$f=\lim\dfrac{n^4-4n^6}{n\left(\sqrt{n^4+1}+\sqrt{4n^6+1}\right)}=\lim\dfrac{\dfrac{1}{n^2}-6}{\sqrt{\dfrac{1}{n^6}+\dfrac{1}{n^{10}}}+\sqrt{\dfrac{4}{n^4}+\dfrac{1}{n^{10}}}}=\dfrac{-6}{0}=-\infty$

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LN

lu nguyễn

11 tháng 1 2020

tìm các giới hạn sau

a,$lim\frac{\left(n^2+1\right)\left(2n+3\right)}{\sqrt{n^4-n^2+1}}$

b, lim$\frac{\left(-3^n-6^n\right)}{\left(-3\right)^{n+1}-5^{n+1}}$

c,lim$\left(\sqrt{n^4+1}+n-1\right)$

d, $\sqrt[3]{1+2n-n^3}$

#Toán lớp 11

1

AH

Akai Haruma

Giáo viên

11 tháng 1 2020

Bạn muốn tìm giới hạn nhưng lại không chỉ rõ $n$ chạy đến đâu?

Điển hình như câu 1:

$n\to 0$ thì giới hạn là $3$

$n\to \pm \infty$ thì giới hạn là $\pm \infty$

Bạn phải ghi rõ đề ra chứ?

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T

títtt

13 tháng 10 2023

1) tính $\lim\limits_{n\rightarrow\infty}\dfrac{6n-8}{n-1}$

2) $\lim\limits_{n\rightarrow\infty}\dfrac{n^2+5n-3}{4n^3-2n+5}$

3) $\lim\limits_{n\rightarrow\infty}\left(-2n^5+4x^4-3n^2+4\right)$

#Toán lớp 11

1

MH

Minh Hiếu

13 tháng 10 2023

1) $\lim\limits_{n\rightarrow\infty}\dfrac{6n-8}{n-1}=\lim\limits_{n\rightarrow\infty}\dfrac{2n\left(1-\dfrac{4}{n}\right)}{n\left(1-\dfrac{1}{n}\right)}=2$

2) $\lim\limits_{n\rightarrow\infty}\dfrac{n^2+5n-3}{4n^3-2n+5}=\lim\limits_{n\rightarrow\infty}\dfrac{n^2\left(1+\dfrac{5}{n}-\dfrac{3}{n^2}\right)}{n^3\left(4-\dfrac{2}{n^2}+\dfrac{5}{n^3}\right)}=\dfrac{1}{4n}=\infty$

3) $\lim\limits_{n\rightarrow\infty}\left(-2n^5+4n^4-3n^2+4\right)=\lim\limits_{n\rightarrow\infty}n^5\left(-2+\dfrac{4}{n}-\dfrac{3}{n^2}+\dfrac{4}{n^5}\right)=-2n^5=-\infty$

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M

Maoromata

19 tháng 1 2021

Tính :6/ lim$\dfrac{-n^2+2n+1}{\sqrt{3n^4+2}}$

7/ lim $\dfrac{\sqrt{n^3-2n+5}}{3+5n}$

10/ lim$\dfrac{1+3+5+...+\left(2n+1\right)}{3n^3+4}$

#Toán lớp 11

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