$limu_n=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)...\left(1-\frac{1}{n^2}\right)\\ n\rightarrow\infty$$limu_n=n\left(\sqrt[3]{n^{\alpha} 1}-n\right)\\ n\rightarrow\infty$\(lim\frac{...

1

NN

Nhan Nguyen

22 tháng 9 2016

thầy cô và các bạn biết câu nào giúp mình câu đó em rất cảm ơn ạ

bài 1 a. $\lim\limits_{n\rightarrow+\infty}\left(\sqrt[3]{n^3+3n+1}-n\right)$ b. $\lim\limits_{n\rightarrow+\infty}\left(\sqrt[3]{n^3+2n}-\sqrt{n^2+1}\right)$ c.$\lim\limits_{n\rightarrow+\infty}\left(\sqrt[3]{n^3+n}-\sqrt[3]{n^3-2n^2}\right)$ bài 2 a. $\lim\limits_{n\rightarrow+\infty}\left(2n-\sqrt{n^2+3n}\right)$ b. $\lim\limits_{n\rightarrow+\infty}\left(\sqrt{n^2+2}-\sqrt{3n+1}\right)$ c....

TP

Trần Phương Thảo

28 tháng 2 2020

2

NV

Nguyễn Việt Lâm

Giáo viên

28 tháng 2 2020

Giới hạn của dãy nên bạn tự hiểu n tiến tới dương vô cực

1.

$lim\frac{3n+1}{\sqrt[3]{\left(n^3+3n+1\right)^2}+n\sqrt{n^3+3n+1}+n^2}=lim\frac{3+\frac{1}{n}}{\sqrt[3]{\frac{\left(n^3+3n+1\right)^2}{n^3}}+\sqrt{n^3+3n+1}+n}=\frac{3}{\infty}=0$

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

$=lim\left(\frac{2}{\sqrt[3]{\frac{\left(n^3+2n\right)^2}{n^3}}+\sqrt[3]{n^3+2n}+n}-\frac{1}{n+\sqrt{n^2+1}}\right)=0-0=0$

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

$=lim\left(\frac{2+\frac{1}{n}}{\sqrt[3]{\left(1+\frac{1}{n^2}\right)^2}+\sqrt[3]{\left(1+\frac{1}{n^2}\right)\left(1-\frac{2}{n}\right)}+\sqrt[3]{\left(1-\frac{2}{n}\right)^2}}\right)=\frac{2}{1+1.1+1}=\frac{2}{3}$

2.

a$=lim\left[n\left(2-\sqrt{1+\frac{3}{n}}\right)\right]=+\infty\left(2-1\right)=+\infty$

$b=lim\left[n\left(\sqrt{1+\frac{2}{n^2}}-\sqrt{\frac{3}{n}+\frac{1}{n^2}}\right)\right]=+\infty\left(1-0\right)=+\infty$

$c=lim\left[n^3\left(\frac{sin2n}{n^2}-3\right)\right]=+\infty\left(0-3\right)=-\infty$

NT

Ngô Tiến Đạt

VIP

9 tháng 8 2022

Jehheheu3uehegayaya

AK

An Khanh Nguyên

4 tháng 3 2020

Cho dãy số $\left(u_n\right)$ : $\left\{{}\begin{matrix}u_1=\frac{5}{2}\\u_{n+1}=\frac{1}{2}u_n^2-u_n+2\end{matrix}\right.$ với n=1,2,3... Chứng minh rằng $\lim\limits_{n\rightarrow+\infty}u_n=+\infty$ và tìm $\lim\limits_{n\rightarrow+\infty}\left(\frac{1}{u_1}+\frac{1}{u_2}+...+\frac{1}{u_n}\right)$ ?

1

NB

Nguyễn Bá Hùng

8 tháng 8 2022

1) Có $u_{n+1}-u_n=\dfrac{1}{2}u^2_n-2u_n+2=\dfrac{1}{2}\left(u_n-2\right)^2$ (1)

+) CM $u_n>2$ (n thuộc N*)

n=1 : u₁= 5/2 > 2 (đúng)

Giả sử n=k, u_k > 2 (k thuộc N*)

Ta cần CM n = k + 1. Thật vậy ta có:

$u_{k+1}=\dfrac{1}{2}u^2_k-u_k+2=\dfrac{1}{2}\left(u_k-2\right)^2+u_k$ (đúng)

Vậy u_n > 2 (n thuộc N*) (2)

Từ (1) (2) => u_n+1- u_n> 0, hay u_n+1> u_n

=> (u_n) là dãy tăng => $\lim\limits_{n\rightarrow\infty}u_n=+\infty$

2) $2u_{n+1}=u^2_n-2u_n+4$

$\Leftrightarrow2u_{n+1}-4=u^2_n-2u_n$

$\Leftrightarrow2\left(u_{n+1}-2\right)=u_n\left(u_n-2\right)$

$\Leftrightarrow\dfrac{1}{u_{n+1}-2}=\dfrac{2}{u_n\left(u_n-2\right)}=\dfrac{1}{u_n-2}-\dfrac{1}{u_n}$

$\Leftrightarrow\dfrac{1}{u_n}=\dfrac{1}{u_n-2}-\dfrac{1}{u_{n+1}-2}$

$S=\dfrac{1}{u_1}+\dfrac{1}{u_2}+...+\dfrac{1}{u_n}$

$=\dfrac{1}{u_1-2}-\dfrac{1}{u_2-2}+\dfrac{1}{u_2-2}+...-\dfrac{1}{u_{n+1}-2}$

$=\dfrac{1}{u_1-2}-\dfrac{1}{u_{n+1}-2}$

$=2-\dfrac{1}{u_{n+1}-2}$

$\Leftrightarrow\lim\limits_{n\rightarrow\infty}S=2$

T

títtt

15 tháng 10 2023

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

2) Tính giới hạn $\lim\limits_{n\rightarrow\infty}\left(\sqrt{4n^2+1}-\sqrt[3]{8n^3+n}\right)$

1

NL

Nguyễn Lê Phước Thịnh

15 tháng 10 2023

1:

$=\lim\limits_{n\rightarrow\infty}\dfrac{n^3+3n^2+1-n^3}{\sqrt[3]{n^3+3n^2+1}+n}$

$=\lim\limits_{n\rightarrow\infty}\dfrac{3n^2+1}{\sqrt[3]{n^3+3n^2+1}+n}$

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

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

$=\lim\limits_{n\rightarrow\infty}n\cdot\lim\limits_{n\rightarrow\infty}\dfrac{3+\dfrac{1}{n^2}}{\sqrt[3]{1+\dfrac{3}{n}+\dfrac{1}{n^3}}+1}$

$=+\infty$ vì $\left\{{}\begin{matrix}\lim\limits_{n\rightarrow\infty}n=+\infty\\\lim\limits_{n\rightarrow\infty}\dfrac{3+\dfrac{1}{n^2}}{\sqrt[3]{1+\dfrac{3}{n}+\dfrac{1}{n^3}}+1}=\dfrac{3}{2}>0\end{matrix}\right.$

2:

$=\lim\limits_{n\rightarrow\infty}\left(\sqrt{4n^2+1}-2n+2n-\sqrt[3]{8n^3+n}\right)$

$=\lim\limits_{n\rightarrow\infty}\dfrac{4n^2+1-4n^2}{\sqrt{4n^2+1}+2n}+\lim\limits_{n\rightarrow\infty}\dfrac{8n^3-8n^3-n}{4n^2+2n\cdot\sqrt[3]{8n^3+n}+\left(\sqrt[3]{8n^3+n}\right)^2}$

$=\lim\limits_{n\rightarrow\infty}\dfrac{1}{\sqrt{4n^2+1}+2n}+\lim\limits_{n\rightarrow\infty}\dfrac{-n}{4n^2+2n\cdot n\cdot\sqrt[3]{8+\dfrac{1}{n^3}}+\left(n\cdot\sqrt[3]{8+\dfrac{1}{n^3}}\right)^2}$

$=\lim\limits_{n\rightarrow\infty}\dfrac{-n}{4n^2+2n^2\cdot\sqrt[3]{8+\dfrac{1}{n^3}}+n^2\cdot\sqrt[3]{\left(8+\dfrac{1}{n^3}\right)^2}}$

$=\lim\limits_{n\rightarrow\infty}\dfrac{-1}{4n+2n\cdot\sqrt[3]{8+\dfrac{1}{n^3}}+n\cdot\sqrt[3]{\left(8+\dfrac{1}{n^3}\right)^2}}$

$=0$

Bài 1 a. $\lim\limits_{x\rightarrow-\infty}\frac{\sqrt{4x^2}+1}{3x-1}$ b. $\lim\limits_{x\rightarrow+\infty}\frac{\sqrt{9x^2+x+1}-\sqrt{4x^2+2x+1}}{x+1}$ c. $\lim\limits_{x\rightarrow+\infty}\frac{\sqrt{x+2x+3}+4x+1}{\sqrt{4x^2+1}+2-x}$ d. $\lim\limits_{x\rightarrow+\infty}\frac{3x-2\sqrt{x}+\sqrt{x^4-5x}}{2x^2+4x-5}$ Bài 2 a. $\lim\limits_{x\rightarrow-\infty}\frac{2x+1}{x-1}$ b. $\lim\limits_{x\rightarrow-\infty}\frac{2x^3+3}{x^3-2x^2+1}$ c....

TP

Trần Phương Thảo

15 tháng 3 2020

2

NV

Nguyễn Việt Lâm

Giáo viên

15 tháng 3 2020

Bài 1:

$a=\lim\limits_{x\rightarrow-\infty}\frac{2\left|x\right|+1}{3x-1}=\lim\limits_{x\rightarrow-\infty}\frac{-2x+1}{3x-1}=\lim\limits_{x\rightarrow-\infty}\frac{-2+\frac{1}{x}}{3-\frac{1}{x}}=-\frac{2}{3}$

$b=\lim\limits_{x\rightarrow+\infty}\frac{\sqrt{9+\frac{1}{x}+\frac{1}{x^2}}-\sqrt{4+\frac{2}{x}+\frac{1}{x^2}}}{1+\frac{1}{x}}=\frac{\sqrt{9}-\sqrt{4}}{1}=1$

$c=\lim\limits_{x\rightarrow+\infty}\frac{\sqrt{1+\frac{2}{x}+\frac{3}{x^2}}+4+\frac{1}{x}}{\sqrt{4+\frac{1}{x^2}}+\frac{2}{x}-1}=\frac{1+4}{\sqrt{4}-1}=5$

$d=\lim\limits_{x\rightarrow+\infty}\frac{\frac{3}{x}-\frac{2}{x\sqrt{x}}+\sqrt{1-\frac{5}{x^3}}}{2+\frac{4}{x}-\frac{5}{x^2}}=\frac{1}{2}$

NV

Nguyễn Việt Lâm

Giáo viên

15 tháng 3 2020

Bài 2:

$a=\lim\limits_{x\rightarrow-\infty}\frac{2+\frac{1}{x}}{1-\frac{1}{x}}=2$

$b=\lim\limits_{x\rightarrow-\infty}\frac{2+\frac{3}{x^3}}{1-\frac{2}{x}+\frac{1}{x^3}}=2$

$c=\lim\limits_{x\rightarrow+\infty}\frac{x^2\left(3+\frac{1}{x^2}\right)x\left(5+\frac{3}{x}\right)}{x^3\left(2-\frac{1}{x^3}\right)x\left(1+\frac{4}{x}\right)}=\frac{15}{+\infty}=0$

T

títtt

14 tháng 10 2023

1) tính giới hạn $\lim\limits_{n\rightarrow\infty}\left(\sqrt{n^2+6n}-n\right)$

2) tính giới hạn $\lim\limits_{n\rightarrow\infty}\left(\sqrt{n+1}-\sqrt{n-1}\right)$

1

AH

Akai Haruma

Giáo viên

14 tháng 10 2023

Lời giải:
1.

$\lim\limits_{n\to \infty}(\sqrt{n^2+6n}-n)=\lim\limits_{n\to \infty}\frac{6n}{\sqrt{n^2+6n}+n}=\lim\limits_{n\to \infty}\frac{6}{\sqrt{1+\frac{6}{n}}+1}=\frac{6}{1+1}=3$

2.

$\lim\limits_{n\to \infty}(\sqrt{n+1}-\sqrt{n-1})=\lim\limits_{n\to \infty}\frac{(n+1)-(n-1)}{\sqrt{n+1}+\sqrt{n-1}}=\lim\limits_{n\to \infty}\frac{2}{\sqrt{n+1}+\sqrt{n-1}}=0$ do $\sqrt{n+1}+\sqrt{n-1}\to \infty$ khi $n\to \infty$

PC

Pé Coldly

30 tháng 11 2023

Tính giới hạn sau:

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

2) $\lim\limits_{n\rightarrow\infty}\dfrac{1}{n}[\left(x+\dfrac{a}{n}\right)+\left(x+\dfrac{2a}{n}\right)+...+\left(x+\dfrac{\left(n-1\right)a}{n}\right)]$

3) $\lim\limits_{n\rightarrow\infty}\dfrac{1^3+2^3+...+n^3}{n^4}$

#Toán lớp 12

3

AH

Akai Haruma

Giáo viên

30 tháng 11 2023

1.

Trước hết bạn nhớ công thức:

$1^2+2^2+....+n^2=\frac{n(n+1)(2n+1)}{6}$ (cách cm ở đây: https://hoc24.vn/cau-hoi/tinh-tongs-122232n2.83618073020)

Áp vào bài:

$\lim\frac{1}{n^3}[1^2+2^2+....+(n-1)^2]=\lim \frac{1}{n^3}.\frac{(n-1)n(2n-1)}{6}=\lim \frac{n(n-1)(2n-1)}{6n^3}$

$=\lim \frac{(n-1)(2n-1)}{6n^2}=\lim (\frac{n-1}{n}.\frac{2n-1}{6n})=\lim (1-\frac{1}{n})(\frac{1}{3}-\frac{1}{6n})$

$=1.\frac{1}{3}=\frac{1}{3}$

AH

Akai Haruma

Giáo viên

30 tháng 11 2023

2.

$\lim \frac{1}{n}\left[(x+\frac{a}{n})+(x+\frac{2a}{n})+...+(x.\frac{(n-1)a}{n}\right]$

$=\lim \frac{1}{n}\left[\underbrace{(x+x+...+x)}_{n-1}+\frac{a(1+2+...+n-1)}{n} \right]$

$=\lim \frac{1}{n}[(n-1)x+a(n-1)]=\lim \frac{n-1}{n}(x+a)=\lim (1-\frac{1}{n})(x+a)$

$=x+a$

T

títtt

15 tháng 10 2023

1) Tính $I=\lim\limits_{n\rightarrow\infty}\left(\sqrt{n^2+2}-\sqrt{n^2-1}\right)$

2) Tính giới hạn $\lim\limits_{n\rightarrow\infty}\left(\sqrt{n^2+2n+2}+n\right)$

1, $\lim\limits_{x\rightarrow1}\frac{2x^2-3x+1}{x^3-x^2-x+1}$ 2, $\lim\limits_{x\rightarrow2}\frac{x-\sqrt{x+2}}{\sqrt{4x+1}-3}$ 3, $\lim\limits_{x\rightarrow0}\frac{1-\sqrt[3]{x-1}}{x}$ 4, $\lim\limits_{x\rightarrow-\infty}\frac{x^2-5x+1}{x^2-2}$ 5, $\lim\limits_{x\rightarrow+\infty}\frac{2x^2-4}{x^3+3x^2-9}$ 6, $\lim\limits_{x\rightarrow2^-}\frac{2x-1}{x-2}$ 7, $\lim\limits_{x\rightarrow3^+}\frac{8+x-x^2}{x-3}$ 8,...

1

NL

Nguyễn Lê Phước Thịnh

15 tháng 10 2023

1: $I=\lim\limits_{n\rightarrow\infty}\dfrac{n^2+2-n^2+1}{\sqrt{n^2+2}+\sqrt{n^2-1}}$

$=\lim\limits_{n\rightarrow\infty}\dfrac{3}{\sqrt{n^2+2}+\sqrt{n^2-1}}$

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

=0

2: $\lim\limits_{n\rightarrow\infty}\sqrt{n^2+2n+2}+n$

$=\lim\limits_{n\rightarrow\infty}\dfrac{n^2+2n+2-n^2}{\sqrt{n^2+2n+2}-n}$

$=\lim\limits_{n\rightarrow\infty}\dfrac{2n+2}{\sqrt{n^2+2n+2}-n}$

$=\lim\limits_{n\rightarrow\infty}\dfrac{2+\dfrac{1}{n}}{\sqrt{1+\dfrac{2}{n}+\dfrac{2}{n^2}}-1}$

$=+\infty$

Đúng(1)

NA

Ngọc Ánh Nguyễn Thị

11 tháng 3 2020

9

AH

Akai Haruma

Giáo viên

12 tháng 3 2020

Bài 2:

$\lim\limits_{x\to 2}\frac{x-\sqrt{x+2}}{\sqrt{4x+1}-3}=\lim\limits_{x\to 2}\frac{x^2-x-2}{(x+\sqrt{x+2}).\frac{4x+1-9}{\sqrt{4x+1}+3}}=\lim\limits_{x\to 2}\frac{(x-2)(x+1)(\sqrt{4x+1}+3)}{(x+\sqrt{x+2}).4(x-2)}=\lim\limits_{x\to 2}\frac{(x+1)(\sqrt{4x+1}+3)}{4(x+\sqrt{x+2})}=\frac{9}{8}$

Bài 3:

$\lim\limits_{x\to 0-}\frac{1-\sqrt[3]{x-1}}{x}=-\infty $

$\lim\limits_{x\to 0+}\frac{1-\sqrt[3]{x-1}}{x}=+\infty $

Bài 4:

$\lim\limits_{x\to -\infty}\frac{x^2-5x+1}{x^2-2}=\lim\limits_{x\to -\infty}\frac{1-\frac{5}{x}+\frac{1}{x^2}}{1-\frac{2}{x^2}}=1$

Bài 5:

$\lim\limits_{x\to +\infty}\frac{2x^2-4}{x^3+3x^2-9}=\lim\limits_{x\to +\infty}\frac{\frac{2}{x}-\frac{4}{x^3}}{1+\frac{3}{x}-\frac{9}{x^3}}=0$

AH

Akai Haruma

Giáo viên

12 tháng 3 2020

Bài 6:

$\lim\limits_{x\to 2- }\frac{2x-1}{x-2}=\lim\limits_{x\to 2-}\frac{2(x-2)+3}{x-2}=\lim\limits_{x\to 2-}\left(2+\frac{3}{x-2}\right)=-\infty $

Bài 7:

$\lim\limits _{x\to 3+ }\frac{8+x-x^2}{x-3}=\lim\limits _{x\to 3+}\frac{1}{x-3}.\lim\limits _{x\to 3+}(8+x-x^2)=2(+\infty)=+\infty $

Bài 8:

$\lim\limits _{x\to -\infty}(8+4x-x^3)=\lim\limits _{x\to -\infty}(-x^3)=+\infty $

Bài 9:

$\lim\limits _{x\to -1}\frac{\sqrt[3]{x}+1}{\sqrt{x^2+3}-2}=\lim\limits _{x\to -1}\frac{x+1}{\sqrt[3]{x^2}-\sqrt[3]{x}+1}.\frac{\sqrt{x^2+3}+2}{x^2+3-4}=\lim\limits _{x\to -1}\frac{x+1}{\sqrt[3]{x^2}-\sqrt[3]{x}+1}.\frac{\sqrt{x^2+3}+2}{(x-1)(x+1)}$

$\lim\limits _{x\to -1}\frac{\sqrt{x^2+3}+2}{(\sqrt[3]{x^2}-\sqrt[3]{x}+1)(x-1)}=\frac{-2}{3}$

T

títtt

15 tháng 10 2023

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

2) Tính giới hạn $\lim\limits_{n\rightarrow\infty}\left(\dfrac{4n-\sqrt{16n^2+1}}{n+1}\right)$

3) Tính giới hạn $\lim\limits_{n\rightarrow\infty}\left(\dfrac{\sqrt{9n^2+n+1}-3n}{2n}\right)$