Bài 1 : tính giới hạn của $\lim\limits_{x\rightarrow1}\frac{4x^6-5x^5 x}{\left(1-x\right)^2}$ Bài 2: chứng minh rằng $\sqrt{x^2 px q}=\left|x \frac{p}{2}\right| \varepsilon\left(x\right)$ với...

3

AH

Akai Haruma

Giáo viên

3 tháng 4 2020

Bài 1:
$\lim\limits _{x\to 1}\frac{4x^6-5x^5+x}{(1-x)^2}=\lim\limits _{x\to 1}\frac{x(x-1)^2(4x^3+3x^2+2x+1)}{(1-x)^2}$

$=\lim\limits _{x\to 1}x(4x^3+3x^2+2x+1)=1(4.1^3+3.1^2+2.1+1)=10$

AH

Akai Haruma

Giáo viên

3 tháng 4 2020

Bài 3:

$\lim\limits _{x\to +\infty}[\sqrt{9x^2-4x+3}-(ax+b)]=0$

$\Rightarrow \lim\limits _{x\to +\infty}\frac{\sqrt{9x^2-4x+3}-(ax+b)}{x}=0$

$\Leftrightarrow \lim\limits _{x\to +\infty}\left(\sqrt{9-\frac{4}{x}+\frac{3}{x^2}}-a+\frac{b}{x}\right)=0$

$\Leftrightarrow a=3$

Thay $a=3$ vào đk ban đầu:

$\lim\limits _{x\to +\infty}[\sqrt{9x^2-4x+3}-3x-b]=0$

$\Leftrightarrow \lim\limits _{x\to +\infty} (\sqrt{9x^2-4x+3}-3x)=b$

$\Leftrightarrow \lim\limits _{x\to +\infty}\frac{-4x+3}{\sqrt{9x^2-4x+3}+3x}=b$

$\Leftrightarrow \lim\limits _{x\to +\infty}\frac{-4+\frac{3}{x}}{\sqrt{9-\frac{4}{x}+\frac{3}{x}}+3}=b$

$\Leftrightarrow \frac{-4}{6}=b\Leftrightarrow b=-\frac{2}{3}$

Bài 1:Cho $\lim\limits_{x\rightarrow1}\dfrac{f\left(x\right)-10}{x-1}=5$ ,$g\left(x\right)=\sqrt{f\left(x\right)+6}-2\sqrt[3]{f\left(x\right)-2}$Tính $\lim\limits_{x\rightarrow1}\dfrac{1}{\left(\sqrt{x}-1\right)g\left(x\right)}$Bài 2: Cho $\lim\limits_{x\rightarrow1}\dfrac{\sqrt{2ax^2+30}-bx-5}{x^3-3x+2}=c\left(a;b;c\in R\right)$Tính giá trị $P=a^2+b^2+36c$Bài 3: Cho a;b là các số nguyên dương....

TM

Trần Minh

14 tháng 5 2021

9

NV

Nguyễn Việt Lâm

Giáo viên

14 tháng 5 2021

Mấy câu này bạn cần giải theo kiểu trắc nghiệm hay tự luận nhỉ?

TM

Trần Minh

14 tháng 5 2021

Em cần kiểu tự luận ạ

Bài 1 a. $\lim\limits_{x\rightarrow-\infty}\left(\sqrt[3]{x^3-x^2}-x\right)$ b. $\lim\limits_{x\rightarrow+\infty}\left(\sqrt[3]{x^3+5x^2}-\sqrt[3]{x^3+8x}\right)$ c. $\lim\limits_{x\rightarrow+\infty}\left(\sqrt[3]{x^3+1}-x\right)$ Bài 2 a. $\lim\limits_{x\rightarrow1^-}\left(\frac{2}{x^2-1}-\frac{1}{x-1}\right)$ b. $\lim\limits_{x\rightarrow1^+}\left(\frac{1}{1-x}-\frac{3}{1-x^3}\right)$ c....

TP

Trần Phương Thảo

15 tháng 3 2020

1

NV

Nguyễn Việt Lâm

Giáo viên

15 tháng 3 2020

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

$b=\lim\limits_{x\rightarrow+\infty}\frac{5x^2-8x}{\sqrt[3]{\left(x^3+5x^2\right)^2}+\sqrt[3]{\left(x^3+5x^2\right)\left(x^3+8x\right)}+\sqrt[3]{\left(x^3+8x\right)^2}}$

$=\lim\limits_{x\rightarrow+\infty}\frac{5-\frac{8}{x}}{\sqrt[3]{\left(1+\frac{5}{x}\right)^2}+\sqrt[3]{\left(1+\frac{5}{x}\right)\left(1+\frac{8}{x^2}\right)}+\sqrt[3]{\left(1+\frac{8}{x^2}\right)^2}}=\frac{5}{3}$

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

Bài 2:

$a=\lim\limits_{x\rightarrow1^-}\left(\frac{1-x}{\left(x-1\right)\left(x+1\right)}\right)=\lim\limits_{x\rightarrow1^-}\frac{-1}{x+1}=-\frac{1}{2}$

$b=\lim\limits_{x\rightarrow1^+}\left(\frac{x^2+x+1-3}{\left(1-x\right)\left(x^2+x+1\right)}\right)=\lim\limits_{x\rightarrow1^+}\frac{\left(x-1\right)\left(x+2\right)}{\left(1-x\right)\left(x^2+x+1\right)}=\lim\limits_{x\rightarrow1^+}\frac{-x-2}{x^2+x+1}=-1$

$c=\lim\limits_{x\rightarrow2^+}\left(\frac{1}{\left(x-1\right)\left(x-2\right)}-\frac{1}{\left(x-2\right)\left(x-3\right)}\right)=\lim\limits_{x\rightarrow2^+}\frac{-2}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}$

Do $x\rightarrow2^+\Rightarrow x>2\Rightarrow x-2>0\Rightarrow\left(x-1\right)\left(x-2\right)\left(x-3\right)\rightarrow0^-$

$\Rightarrow\lim\limits_{x\rightarrow2^+}\frac{-2}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}=+\infty$

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,...

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:

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

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

Bài 1 a. $\lim\limits_{x\rightarrow-1}\frac{x^5+1}{x^3+1}$ b. $\lim\limits_{x\rightarrow1}\frac{4x^6-5x^5+x}{\left(1-x\right)^2}$ c. $\lim\limits_{x\rightarrow0}\frac{\left(1+x\right)\left(1+2x\right)\left(1+3x\right)-1}{x}$ d. $\lim\limits_{x\rightarrow0}\frac{\left(1+x\right)^5-\left(1+5x\right)}{x^5+x^2}$ Bài 2 a. $\lim\limits_{x\rightarrow1}\frac{x^m-1}{x^n-1}$ b. $\lim\limits_{x\rightarrow a}\frac{x-a}{x^n-a^n}\left(n\in Z^+,a\ne0\right)$ ...

TP

Trần Phương Thảo

1 tháng 4 2020

1

NV

Nguyễn Việt Lâm

Giáo viên

3 tháng 4 2020

Bài 1:

a. $\lim\limits_{x\rightarrow-1}\frac{x^5+1}{x^3+1}=\lim\limits_{x\rightarrow-1}\frac{5x^4}{3x^2}=\frac{5}{3}$

b. $\lim\limits_{x\rightarrow1}\frac{4x^6-5x^5+x}{\left(x-1\right)^2}=\lim\limits_{x\rightarrow1}\frac{24x^5-25x^4+1}{2\left(x-1\right)}=\lim\limits_{x\rightarrow1}\frac{120x^4-100x^3}{2}=\frac{120-100}{2}=10$

c. $\lim\limits_{x\rightarrow0}\frac{\left(1+2x\right)\left(1+3x\right)x}{x}+\lim\limits_{x\rightarrow0}\frac{\left(1+3x\right)2x}{x}+\lim\limits_{x\rightarrow0}\frac{3x+1-1}{x}=1+2+3=6$

d. $\lim\limits_{x\rightarrow0}\frac{\left(1+x\right)^5-\left(1+5x\right)}{x^5+x^2}=\lim\limits_{x\rightarrow0}\frac{5\left(1+x\right)^4-5}{5x^4+2x}$

$=\lim\limits_{x\rightarrow0}\frac{20\left(1+x\right)^3}{20x^3+2}=\frac{20}{2}=10$

Bài 2:

$\lim\limits_{x\rightarrow1}\frac{x^m-1}{x^n-1}=\lim\limits_{x\rightarrow1}\frac{mx^{m-1}}{nx^{n-1}}=\frac{m}{n}$

$\lim\limits_{x\rightarrow a}\frac{x-a}{x^n-a^n}=\lim\limits_{x\rightarrow a}\frac{1}{nx^{n-1}}=\frac{1}{n.a^{n-1}}$

QT

Quoc Tran Anh Le

Giáo viên

22 tháng 9 2023

Tính các giới hạn sau:

a) $\mathop {\lim }\limits_{x \to 2} \left( {{x^2} - 4x + 3} \right);$

b) $\mathop {\lim }\limits_{x \to 3} \frac{{{x^2} - 5x + 6}}{{x - 3}};$

c) $\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt x - 1}}{{x - 1}}.$

1

HQ

Hà Quang Minh

Giáo viên

22 tháng 9 2023

a) $\mathop {\lim }\limits_{x \to 2} \left( {{x^2} - 4x + 3} \right) = \mathop {\lim }\limits_{x \to 2} {x^2} - \mathop {\lim }\limits_{x \to 2} \left( {4x} \right) + 3 = {2^2} - 4.2 + 3 = - 1$

b) $\mathop {\lim }\limits_{x \to 3} \frac{{{x^2} - 5x + 6}}{{x - 3}} = \mathop {\lim }\limits_{x \to 3} \frac{{\left( {x - 3} \right)\left( {x - 2} \right)}}{{x - 3}} = \mathop {\lim }\limits_{x \to 3} \left( {x - 2} \right) = \mathop {\lim }\limits_{x \to 3} x - 2 = 3 - 2 = 1$

c) $\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt x - 1}}{{x - 1}} = \mathop {\lim }\limits_{x \to 1} \frac{{\sqrt x - 1}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}} = \mathop {\lim }\limits_{x \to 1} \frac{1}{{\sqrt x + 1}} = \frac{1}{{\sqrt 1 + 1}} = \frac{1}{2}$

a. $\lim\limits_{x\rightarrow a}\frac{x\sqrt{x}-a\sqrt{a}}{\sqrt{x}-\sqrt{a}}$ e. $\lim\limits_{x\rightarrow0}\frac{\sqrt{1+x}-\sqrt[3]{1+x}}{x}$ b. $\lim\limits_{x\rightarrow1}\frac{\sqrt[n]{x}-1}{\sqrt[m]{x}-1}\left(m,n\in Z^+\right)$ f. $\lim\limits_{x\rightarrow2}\frac{\sqrt[3]{8x+11}-\sqrt{x+7}}{x^2-3x+2}$ c. $\lim\limits_{x\rightarrow1}\frac{\left(1-\sqrt{x}\right)\left(1-\sqrt[3]{x}\right)\left(1-\sqrt[4]{x}\right)\left(1-\sqrt[5]{x}\right)}{\left(1-x\right)^4}$ g....

TP

Trần Phương Thảo

15 tháng 3 2020

2

NV

Nguyễn Việt Lâm

Giáo viên

15 tháng 3 2020

$a=\lim\limits_{x\rightarrow a}\frac{\left(\sqrt{x}-\sqrt{a}\right)\left(x+\sqrt{ax}+a\right)}{\sqrt{x}-\sqrt{a}}=\lim\limits_{x\rightarrow a}\left(x+\sqrt{ax}+a\right)=3a$

$b=\lim\limits_{x\rightarrow1}\frac{x^{\frac{1}{n}}-1}{x^{\frac{1}{m}}-1}=\lim\limits_{x\rightarrow1}\frac{\frac{1}{n}x^{\frac{1-n}{n}}}{\frac{1}{m}x^{\frac{1-m}{m}}}=\frac{\frac{1}{n}}{\frac{1}{m}}=\frac{m}{n}$

Ta có:

$\lim\limits_{x\rightarrow1}\frac{1-\sqrt[n]{x}}{1-x}=\lim\limits_{x\rightarrow1}\frac{1-x^{\frac{1}{n}}}{1-x}=\lim\limits_{x\rightarrow1}\frac{-\frac{1}{n}x^{\frac{1-n}{n}}}{-1}=\frac{1}{n}$

$\Rightarrow c=\lim\limits_{x\rightarrow1}\frac{\left(1-\sqrt{x}\right)}{1-x}.\frac{\left(1-\sqrt[3]{x}\right)}{\left(1-x\right)}.\frac{\left(1-\sqrt[4]{x}\right)}{\left(1-x\right)}.\frac{\left(1-\sqrt[5]{x}\right)}{\left(1-x\right)}=\frac{1}{2}.\frac{1}{3}.\frac{1}{4}.\frac{1}{5}=\frac{1}{120}$

$d=\lim\limits_{x\rightarrow+\infty}\frac{\sqrt{x+\sqrt{x}}}{\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}}=\lim\limits_{x\rightarrow+\infty}\frac{\sqrt{1+\frac{1}{\sqrt{x}}}}{\sqrt{1+\sqrt{\frac{1}{x}+\frac{1}{x\sqrt{x}}}}+1}=\frac{1}{2}$

NV

Nguyễn Việt Lâm

Giáo viên

15 tháng 3 2020

$e=\lim\limits_{x\rightarrow0}\frac{\sqrt{1+x}-1+1-\sqrt[3]{1+x}}{x}=\lim\limits_{x\rightarrow0}\frac{\frac{x}{\sqrt{1+x}+1}+\frac{x}{1+\sqrt[3]{1+x}+\sqrt[3]{\left(1+x\right)^2}}}{x}$

$=\lim\limits_{x\rightarrow0}\left(\frac{1}{\sqrt{1+x}+1}+\frac{1}{1+\sqrt[3]{1+x}+\sqrt[3]{\left(1+x\right)^2}}\right)=\frac{1}{2}+\frac{1}{3}=\frac{5}{6}$

$f=\lim\limits_{x\rightarrow2}\frac{\sqrt[3]{8x+11}-3+3-\sqrt{x+7}}{\left(x-1\right)\left(x-2\right)}=\lim\limits_{x\rightarrow2}\frac{\frac{8\left(x-2\right)}{\sqrt[3]{\left(8x+11\right)^2}+3\sqrt[3]{8x+11}+9}-\frac{x-2}{3+\sqrt{x+7}}}{\left(x-1\right)\left(x-2\right)}$

$=\lim\limits_{x\rightarrow2}\frac{\frac{8}{\sqrt[3]{\left(8x+11\right)^2}+3\sqrt[3]{8x+11}+9}-\frac{1}{3+\sqrt{x+7}}}{x-1}=\frac{8}{27}-\frac{1}{6}=\frac{7}{54}$

$g=\lim\limits_{x\rightarrow1}\frac{\sqrt[3]{3x-2}-1+1-\sqrt{2x-1}}{\left(x-1\right)\left(x^2+x+1\right)}=\lim\limits_{x\rightarrow1}\frac{\frac{3\left(x-1\right)}{\sqrt[3]{\left(3x-2\right)^2}+\sqrt[3]{3x-2}+1}-\frac{2\left(x-1\right)}{1+\sqrt{2x-1}}}{\left(x-1\right)\left(x^2+x+1\right)}$

$=\lim\limits_{x\rightarrow1}\frac{\frac{3}{\sqrt[3]{\left(3x-2\right)^2}+\sqrt[3]{3x-2}+1}-\frac{2}{1+\sqrt{2x-1}}}{x^2+x+1}=0$

$h=\lim\limits_{x\rightarrow1}\frac{\sqrt[3]{x+9}+\sqrt[3]{2x-6}}{x^3+1}=\frac{\sqrt[3]{10}-\sqrt[3]{4}}{2}$

B

B.Trâm

5 tháng 2 2021

Tính các giới hạn sau:$I_1=\lim\limits_{x\rightarrow1}\dfrac{\left(1-\sqrt{x}\right)\left(1-\sqrt[3]{x}\right)....\left(1-\sqrt[n]{x}\right)}{\left(1-x\right)^{n-1}}$

$I_2=\lim\limits_{x\rightarrow0}\dfrac{\left(\sqrt{1+x^2}+x\right)^n-\left(\sqrt{1+x^2}-x\right)^n}{x}$

Bài 1 a. $\lim\limits_{x\rightarrow-1}\frac{x^5+1}{x^3+1}$ b. $\lim\limits_{x\rightarrow1}\frac{x^6-5x^5+x}{\left(1-x\right)^2}$ c. $\lim\limits_{x\rightarrow0}\frac{\left(1+x\right)\left(1+2x\right)\left(1+3x\right)-1}{x}$ d. $\lim\limits_{x\rightarrow0}\frac{\left(1+x\right)^5-\left(1+x\right)}{x^5+x^2}$ Bài 2 a. $\lim\limits_{x\rightarrow1}\frac{x^m-1}{x^n-1}$ b. $\lim\limits_{x\rightarrow a}\frac{x-a}{x^n-a^n}\left(n\in Z^+,a\ne0\right)$ c....

1

NV

Nguyễn Việt Lâm

Giáo viên

5 tháng 2 2021

Chúng ta tính giới hạn sau:

$\lim\limits_{x\rightarrow1}\dfrac{1-\sqrt[n]{x}}{1-x}$

Cách đơn giản nhất là sử dụng L'Hopital:

$\lim\limits_{x\rightarrow1}\dfrac{1-x^{\dfrac{1}{n}}}{1-x}=\lim\limits_{x\rightarrow1}\dfrac{-\dfrac{1}{n}x^{\dfrac{1}{n}-1}}{-1}=\dfrac{1}{n}$

Phức tạp hơn thì tách mẫu theo hằng đẳng thức

$=\lim\limits_{x\rightarrow1}\dfrac{1-\sqrt[x]{n}}{\left(1-\sqrt[n]{x}\right)\left(1+\sqrt[n]{x}+\sqrt[n]{x^2}+...+\sqrt[n]{x^{n-1}}\right)}=\lim\limits_{x\rightarrow1}\dfrac{1}{1+\sqrt[n]{x}+\sqrt[n]{x^2}+...+\sqrt[n]{x^{n-1}}}=\dfrac{1}{n}$

Tóm lại ta có:

$\lim\limits_{x\rightarrow1}\dfrac{1-\sqrt[n]{x}}{1-x}=\dfrac{1}{n}$

Do đó:

$I_1=\lim\limits_{x\rightarrow1}\left(\dfrac{1-\sqrt[2]{x}}{1-x}\right)\left(\dfrac{1-\sqrt[3]{x}}{1-x}\right)...\left(\dfrac{1-\sqrt[n]{x}}{1-x}\right)=\dfrac{1}{2}.\dfrac{1}{3}...\dfrac{1}{n}=\dfrac{1}{n!}$

Câu 2 cũng vậy: L'Hopital hoặc tách hằng đẳng thức trâu bò (thôi L'Hopital đi cho đỡ sợ)

$I_2=\lim\limits_{x\rightarrow0}\dfrac{\left(\sqrt{1+x^2}+x\right)^n-\left(\sqrt{1+x^2}-x\right)^n}{x}$

$=\lim\limits_{x\rightarrow0}\dfrac{n\left(\sqrt{1+x^2}+x\right)^{n-1}\left(\dfrac{x}{\sqrt{1+x^2}}+1\right)-n\left(\sqrt{1+x^2}-x\right)^{n-1}\left(\dfrac{x}{\sqrt{1+x^2}}-1\right)}{1}$

$=\dfrac{n.1-n\left(-1\right)}{1}=2n$

Đúng(1)

TP

Trần Phương Thảo

15 tháng 3 2020

2

NV

Nguyễn Việt Lâm

Giáo viên

15 tháng 3 2020

Bài 1:

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

$b=\frac{1-5+1}{0}=\frac{-3}{0}=-\infty$

$c=\lim\limits_{x\rightarrow1}\frac{x\left(1+2x\right)\left(1+3x\right)+2x\left(1+3x\right)+3x}{x}=\lim\limits_{x\rightarrow1}\left[\left(1+2x\right)\left(1+3x\right)+2\left(1+3x\right)+3\right]=1+2+3=6$

$d=\lim\limits_{x\rightarrow0}\frac{5\left(1+x\right)^4-1}{5x^4+2x}=\frac{4}{0}=+\infty$

NV

Nguyễn Việt Lâm

Giáo viên

15 tháng 3 2020

Bài 2:

$a=\lim\limits_{x\rightarrow1}\frac{x^m-1}{x^n-1}=\lim\limits_{x\rightarrow1}\frac{mx^{m-1}}{nx^{n-1}}=\frac{m}{n}$

$b=\lim\limits_{x\rightarrow a}\frac{x-a}{x^n-a^n}=\lim\limits_{x\rightarrow a}\frac{1}{nx^{n-1}}=\frac{1}{n.a^{n-1}}$

$c=\lim\limits_{x\rightarrow0}\frac{x+x^2+...+x^n-n}{x-1}=\frac{-n}{-1}=n$

$\left(1+x\right)\left(1+2x\right)...\left(1+nx\right)=x\left(1+2x\right)...\left(1+nx\right)+\left(1+2x\right)\left(1+3x\right)...\left(1+nx\right)$

$=x\left(1+2x\right)...\left(1+nx\right)+2x\left(1+3x\right)...\left(1+nx\right)+\left(1+3x\right)...\left(1+nx\right)$

$=...$

$=x\left(1+2x\right)...\left(1+nx\right)+2x\left(1+3x\right)...\left(1+nx\right)+...+nx+1$

$\Rightarrow\lim\limits_{x\rightarrow0}\frac{\left(1+2x\right)\left(1+3x\right)...\left(1+nx\right)-1}{x}$

$=\lim\limits_{x\rightarrow0}\frac{x\left(1+2x\right)...\left(1+nx\right)+2x\left(1+3x\right)...\left(1+nx\right)+...+nx}{x}$

$=\lim\limits_{x\rightarrow0}\left[\left(1+2x\right)...\left(1+nx\right)+2\left(1+3x\right)...\left(1+nx\right)+...+n\right]$

$=1+2+3+...+n=\frac{n\left(n+1\right)}{2}$

tính các giới hạn sau: ( mình đang tự học bài này nên cần mọi người trình bày chi tiết hộ mình nhé) a;$\lim\limits_{x\rightarrow\left(-3\right)^+}\frac{3x+9}{\left|3+x\right|}$ $\lim\limits_{x\rightarrow\left(-3\right)^+}\frac{3x+9}{\left|3+x\right|}$ b; $\lim\limits_{x\rightarrow0^+}\frac{\sqrt{x}-3x}{4x-2\sqrt{x}}$ c; $\lim\limits_{x\rightarrow1^-}\frac{\sqrt{1-x}}{-x^2-3x+4}$ d;...

LN

lu nguyễn

15 tháng 2 2020

1

NV

Nguyễn Việt Lâm

Giáo viên

15 tháng 2 2020

a/ Do $x\rightarrow-3^+$ nên $x>-3\Rightarrow x+3>0\Rightarrow\left|x+3\right|=x+3$

$\Rightarrow\lim\limits_{x\rightarrow-3^+}\frac{3x+9}{\left|x+3\right|}=\lim\limits_{x\rightarrow-3^+}\frac{3\left(x+3\right)}{x+3}=3$

b/ $=\lim\limits_{x\rightarrow0^+}\frac{\sqrt{x}\left(1-3\sqrt{x}\right)}{\sqrt{x}\left(4\sqrt{x}-2\right)}=\lim\limits_{x\rightarrow0^+}\frac{1-3\sqrt{x}}{4\sqrt{x}-2}=-\frac{1}{2}$

Ở câu này $x\rightarrow0^+$ có nghĩa $x>0$, nó chỉ để căn thức xác định, ngoài ra ko có gì đặc biệt hết

c/ Tương tự câu c, cũng chỉ để căn thức xác định $\left(x< 1\right)$

$\lim\limits_{x\rightarrow1^-}\frac{\sqrt{1-x}}{\left(1-x\right)\left(x+4\right)}=\lim\limits_{x\rightarrow1^-}\frac{1}{\sqrt{1-x}\left(x+4\right)}=+\infty$

d/ Chắc bạn ghi nhầm đề, đây ko phải giới hạn dạng vô định (vì tử khác 0, mẫu bằng 0):

$x\rightarrow\sqrt{2}^-\Rightarrow x< \sqrt{2}\Rightarrow x^4-4< 0$

$\Rightarrow\lim\limits_{x\rightarrow\sqrt{2}^-}\frac{\left|x-2\right|}{x^4-4}=-\infty$