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NV
4 tháng 6 2020

Câu 2:

\(f'\left(x\right)=\frac{-3}{\left(2x-1\right)^2}\)

a/ \(x_0=-1\Rightarrow\left\{{}\begin{matrix}f'\left(x_0\right)=-\frac{1}{3}\\f\left(x_0\right)=0\end{matrix}\right.\)

Pttt: \(y=-\frac{1}{3}\left(x+1\right)=-\frac{1}{3}x-\frac{1}{3}\)

b/ \(y_0=1\Rightarrow\frac{x_0+1}{2x_0-1}=1\Leftrightarrow x_0+1=2x_0-1\Rightarrow x_0=2\)

\(\Rightarrow f'\left(x_0\right)=-\frac{1}{3}\)

Pttt: \(y=-\frac{1}{3}\left(x-2\right)+1\)

c/ \(x_0=0\Rightarrow\left\{{}\begin{matrix}f'\left(x_0\right)=-3\\y_0=-1\end{matrix}\right.\)

Pttt: \(y=-3x-1\)

d/ \(6x+2y-1=0\Leftrightarrow y=-3x+\frac{1}{2}\)

Tiếp tuyến song song d \(\Rightarrow\) có hệ số góc bằng -3

\(\Rightarrow\frac{-3}{\left(2x_0-1\right)^2}=-3\Rightarrow\left(2x_0-1\right)^2=1\Rightarrow\left[{}\begin{matrix}x_0=0\Rightarrow y_0=-1\\x_0=1\Rightarrow y_0=2\end{matrix}\right.\)

Có 2 tiếp tuyến thỏa mãn: \(\left[{}\begin{matrix}y=-3x-1\\y=-3\left(x-1\right)+2\end{matrix}\right.\)

NV
4 tháng 6 2020

Làm câu 1,3 trước, câu 2 hơi dài tối rảnh làm sau:

1/ \(\lim\limits\frac{n^2+2n+1}{2n^2-1}=lim\frac{1+\frac{2}{n}+\frac{1}{n^2}}{2-\frac{1}{n^2}}=\frac{1}{2}\)

\(\lim\limits_{x\rightarrow0}\frac{2\sqrt{x+1}-x^2+2x+2}{x}=\frac{2-0+0+2}{0}=\frac{4}{0}=+\infty\)

Chắc bạn ghi nhầm đề, câu này biểu thức tử số là \(...-x^2+2x-2\) thì hợp lý hơn

3/ \(y'=2sin2x.\left(sin2x\right)'=4sin2x.cos2x=2sin4x\)

b/ \(y'=4x^3-4x\)

c/ \(y'=\frac{3\left(x+2\right)-1\left(3x-1\right)}{\left(x+2\right)^2}=\frac{7}{\left(x+2\right)^2}\)

d/ \(y'=10\left(x^2+x+1\right)^9\left(x^2+x+1\right)'=10\left(x^2+x+1\right)^9.\left(2x+1\right)\)

e/ \(y'=\frac{\left(2x^2-x+3\right)'}{2\sqrt{2x^2-x+3}}=\frac{4x-1}{2\sqrt{2x^2-x+3}}\)

NV
3 tháng 5 2021

1.

\(\lim\limits_{x\rightarrow-1}\dfrac{2x^2-x-3}{x^2-1}=\lim\limits_{x\rightarrow-1}\dfrac{\left(x+1\right)\left(2x-3\right)}{\left(x+1\right)\left(x-1\right)}=\lim\limits_{x\rightarrow-1}\dfrac{2x-3}{x-1}=\dfrac{5}{2}\)

2.

a. \(y'=6x^2-sinx-\dfrac{1}{2\sqrt{x}}\)

b. \(y'=10\left(x^2-5\right)^9.\left(x^2-5\right)'=20x\left(x^2-5\right)^9\)

3.

\(y'=-2x\)

\(k=4\Rightarrow-2x=4\Rightarrow x=-2\Rightarrow y\left(-2\right)=-24\)

Phương trình tiếp tuyến:

\(y=4\left(x+2\right)-24\Leftrightarrow y=4x-16\)

2: \(=lim\left(\dfrac{4n^2+2n+1-4n^2}{\sqrt{4n^2+2n+1}+2n}+2020\right)\)

\(=lim\left(\dfrac{2n+1}{\sqrt{4n^2+2n+1}+2n}+2020\right)\)

\(=lim\left(\dfrac{2+\dfrac{1}{n}}{\sqrt{4+\dfrac{2}{n}+\dfrac{1}{n^2}}+2}+2020\right)\)

\(=\dfrac{2}{2+2}+2020=\dfrac{2}{4}+2020=2020.5\)

 

18 tháng 2 2023

đáp án là 4041/2 chứ đk bạn
 

NV
28 tháng 4 2019

a/ Không phải dạng vô định thì cứ thay trực tiếp vào thôi

\(\lim\limits_{x\rightarrow2}\left(\frac{\sqrt{x^2+60}-2x^2}{x^2-1}\right)=\frac{\sqrt{2^2+60}-2.2^2}{2^2-1}=0\)

b/ Bạn có viết nhầm mẫu số ko? Đề bài thế này hoàn toàn ko chặt chẽ

Số hạng tổng quát \(\frac{1}{4n^2}\) đâu có đúng với 2 số hạng đầu trong dãy?

Dù sao thì, nếu tử số và mẫu số có cùng số số hạng là \(2n\) thì vẫn tính được dựa vào giới hạn kẹp

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

\(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{4n^2}< 1+1+1+...+1=2n\)

\(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2n^2}>\frac{1}{2n^2}+\frac{1}{2n^2}+\frac{1}{2n^2}+...+\frac{1}{2n^2}=2n.\frac{1}{2n^2}=\frac{1}{n}\)

\(\Rightarrow lim\left(\frac{2n\left(2n+1\right)}{2.2n}\right)< lim\left(\frac{1+2+3+...+2n}{1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{4n^2}}\right)< lim\left(\frac{2n\left(2n+1\right)}{\frac{1}{n}}\right)\)

\(lim\left(\frac{2n\left(2n+1\right)}{2.2n}\right)=lim\left(n+\frac{1}{2}\right)=+\infty\)

\(lim\left(\frac{2n\left(2n+1\right)}{\frac{1}{n}}\right)=lim\left(2n^2\left(2n+1\right)\right)=+\infty\)

\(\Rightarrow lim\left(\frac{1+2+3+...+2n}{1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{4n^2}}\right)=+\infty\)

29 tháng 4 2019

Em cảm ơn ạ.

NV
3 tháng 4 2020

Vậy nó ko phải dạng vô định, cứ thay số trực tiếp

\(=\frac{2}{0}=+\infty\)

Nếu là mũ 3 thì nó là dạng 0/0 rút gọn được. Nên chắc là đề ghi nhầm đấy

NV
3 tháng 4 2020

Sry mình ko nhớ 1 chữ về vật lý luôn :<

NV
15 tháng 3 2020

Bài 1:

\(a=\lim\limits_{x\rightarrow+\infty}\frac{\frac{1}{x}+\frac{2}{\sqrt{x}}-1}{1+\frac{3}{x}}=-1\)

\(b=\lim\limits_{x\rightarrow+\infty}\frac{1+\frac{3}{x^2}-\frac{1}{x^3}}{\frac{1}{\sqrt{x}}+\frac{1}{x^2}}=\frac{1}{0}=+\infty\)

\(c=\lim\limits_{x\rightarrow-\infty}\frac{1-2\sqrt{\frac{1}{x^2}-\frac{1}{x}}}{\frac{1}{x}-1}=\frac{1}{-1}=-1\)

Bài 2:

\(a=\lim\limits_{x\rightarrow0}\frac{1-cosx}{1-cos3x}=\lim\limits_{x\rightarrow0}\frac{sinx}{3sin3x}=\lim\limits_{x\rightarrow0}\frac{\frac{sinx}{x}}{9.\frac{sin3x}{3x}}=\frac{1}{9}\)

\(b=\lim\limits_{x\rightarrow0}\frac{cotx-sinx}{x^3}=\frac{\infty}{0}=+\infty\)

\(c=\lim\limits_{x\rightarrow\infty}\frac{sinx}{2x}\)

\(\left|sinx\right|\le1\Rightarrow\left|\frac{sinx}{2x}\right|\le\frac{1}{\left|2x\right|}\)

\(\lim\limits_{x\rightarrow\infty}\frac{1}{2\left|x\right|}=0\Rightarrow\lim\limits_{x\rightarrow\infty}\frac{sinx}{2x}=0\)

AH
Akai Haruma
Giáo viên
25 tháng 1 2020

Lời giải:
a)

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

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

b)

\(\lim\limits_{x\to1}\frac{\sqrt[3]{x^2}-2\sqrt[3]{x}+1}{\left(x-1\right)^2}=\lim\limits_{x\to1}\frac{\left(\sqrt[3]{x}-1\right)^2}{\left(x-1\right)^2}=\lim\limits_{x\to1}\frac{\left(x-1\right)^2}{\left(\sqrt[3]{x^2}+\sqrt[3]{x}+1\right)^2\left(x-1\right)^2}\)

\(=\lim\limits_{x\to1}\frac{1}{\left(\sqrt[3]{x^2}+\sqrt[3]{x}+1\right)^2}=\frac{1}{\left(1+1+1\right)^2}=\frac{1}{9}\)

c)

\(\lim_{x\to 1}\frac{\sqrt[4]{x}-1}{x^3+x^2-2}=\lim_{x\to 1}\frac{\sqrt[4]{x}-1}{(x-1)(x^2+2x+2)}=\lim_{x\to 1}\frac{x-1}{(\sqrt{x}+1)(\sqrt[4]{x}+1)(x-1)(x^2+2x+2)}\)

\(=\lim_{x\to 1}\frac{1}{(\sqrt{x}+1)(\sqrt[4]{x}+1)(x^2+2x+2)}=\frac{1}{(1+1)(1+1)(1+2.1+2)}=\frac{1}{20}\)

d)

\(\lim_{x\to -2}\frac{\sqrt[3]{2x+12}+x}{x^2+2x}=\lim_{x\to -2}\frac{2x+12+x^3}{(\sqrt[3]{(2x+12)^2}-x\sqrt[3]{2x+12}+x^2).x(x+2)}\)

\(=\lim_{x\to -2}\frac{(x+2)(x^2-2x+6)}{(\sqrt[3]{(2x+12)^2}-x\sqrt[3]{2x+12}+x^2).x(x+2)}=\lim_{x\to -2}\frac{x^2-2x+6}{(\sqrt[3]{(2x+12)^2}-x\sqrt[3]{2x+12}+x^2).x}\)

\(=\frac{-7}{12}\)

AH
Akai Haruma
Giáo viên
7 tháng 1 2020

Lời giải:
a)

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

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

b)

\(\lim\limits_{x\to1}\frac{\sqrt[3]{x^2}-2\sqrt[3]{x}+1}{\left(x-1\right)^2}=\lim\limits_{x\to1}\frac{\left(\sqrt[3]{x}-1\right)^2}{\left(x-1\right)^2}=\lim\limits_{x\to1}\frac{\left(x-1\right)^2}{\left(\sqrt[3]{x^2}+\sqrt[3]{x}+1\right)^2\left(x-1\right)^2}\)

\(=\lim\limits_{x\to1}\frac{1}{\left(\sqrt[3]{x^2}+\sqrt[3]{x}+1\right)^2}=\frac{1}{\left(1+1+1\right)^2}=\frac{1}{9}\)

c)

\(\lim_{x\to 1}\frac{\sqrt[4]{x}-1}{x^3+x^2-2}=\lim_{x\to 1}\frac{\sqrt[4]{x}-1}{(x-1)(x^2+2x+2)}=\lim_{x\to 1}\frac{x-1}{(\sqrt{x}+1)(\sqrt[4]{x}+1)(x-1)(x^2+2x+2)}\)

\(=\lim_{x\to 1}\frac{1}{(\sqrt{x}+1)(\sqrt[4]{x}+1)(x^2+2x+2)}=\frac{1}{(1+1)(1+1)(1+2.1+2)}=\frac{1}{20}\)

d)

\(\lim_{x\to -2}\frac{\sqrt[3]{2x+12}+x}{x^2+2x}=\lim_{x\to -2}\frac{2x+12+x^3}{(\sqrt[3]{(2x+12)^2}-x\sqrt[3]{2x+12}+x^2).x(x+2)}\)

\(=\lim_{x\to -2}\frac{(x+2)(x^2-2x+6)}{(\sqrt[3]{(2x+12)^2}-x\sqrt[3]{2x+12}+x^2).x(x+2)}=\lim_{x\to -2}\frac{x^2-2x+6}{(\sqrt[3]{(2x+12)^2}-x\sqrt[3]{2x+12}+x^2).x}\)

\(=\frac{-7}{12}\)

4 tháng 6 2021

\(lim_{x\rightarrow1}\frac{x^3+2x-3}{x^2-x}\)   

\(=lim_{x\rightarrow1}\frac{\left(x-1\right)\left(x^2+x+3\right)}{x\left(x-1\right)}\)   

\(=lim_{x\rightarrow1}\frac{x^2+x+3}{x}\)   

\(=\frac{1^2+1+3}{1}\)   

\(=5\)   

\(lim_{x\rightarrow1}\frac{\sqrt{2x+2}-\sqrt{3x+1}}{x-1}\)   

\(=lim_{x\rightarrow1}\frac{\left(2x+2\right)-\left(3x+1\right)}{\left(x-1\right)\left(\sqrt{2x+2}+\sqrt{3x+1}\right)}\)   

\(=lim_{x\rightarrow1}\frac{2x+2-3x-1}{\left(x-1\right)\left(\sqrt{2x+2}+\sqrt{3x+1}\right)}\)   

\(=lim_{x\rightarrow1}\frac{-x+1}{\left(x-1\right)\left(\sqrt{2x+2}+\sqrt{3x+1}\right)}\)   

\(=lim_{x\rightarrow1}\frac{-1\left(x-1\right)}{\left(x-1\right)\left(\sqrt{2x+2}+\sqrt{3x+1}\right)}\)   

\(=lim_{x\rightarrow1}\frac{-1}{\left(\sqrt{2x+2}+\sqrt{3x+1}\right)}\)   

\(=\frac{-1}{\sqrt{2\cdot1+2}+\sqrt{3\cdot1+1}}\)   

\(=\frac{-1}{2+2}=\frac{-1}{4}\)

20 tháng 4 2022

Võ Ngọc Tú Uyên-41loading...    

NV
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
15 tháng 3 2020

Bài 2:

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

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

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

NV
13 tháng 5 2020

\(L_1=\lim\limits_{x\rightarrow0}\frac{x\left(x^2+3x-2\right)}{x\left(x^4+4\right)}=\lim\limits_{x\rightarrow0}\frac{x^2+3x-2}{x^4+4}=-\frac{1}{2}\)

\(L_2=\lim\limits_{x\rightarrow+\infty}\frac{1-\frac{3}{x^2}+\frac{2}{x^3}}{\left(\frac{4}{x}-2\right)^3}=\frac{1}{\left(-2\right)^3}=-\frac{1}{8}\)

\(L_3=\lim\limits_{x\rightarrow-1}\frac{\left(2x+1\right)\left(x+1\right)}{x\left(x+1\right)}=\lim\limits_{x\rightarrow-1}\frac{2x+1}{x}=1\)

\(L_4=\lim\limits_{x\rightarrow2}\frac{x^2-4x+1}{4-x^2}=\frac{1}{0}=+\infty\)

\(L_5=\lim\limits_{x\rightarrow3}\frac{\sqrt{x+1}-2}{x-2}=\frac{0}{1}=0\)

\(L_6=\lim\limits_{x\rightarrow1}\frac{x+3-\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)\left(\sqrt{x+3}+x+1\right)}=\lim\limits_{x\rightarrow1}\frac{-\left(x-1\right)\left(x+2\right)}{\left(x-1\right)\left(x+1\right)\left(\sqrt{x+3}+x+1\right)}\)

\(=\lim\limits_{x\rightarrow1}\frac{-\left(x+2\right)}{\left(x+1\right)\left(\sqrt{x+3}+x+1\right)}=\frac{-3}{2.4}=-\frac{3}{8}\)

\(L_7=\lim\limits_{x\rightarrow+\infty}\frac{x^2+x+1-\left(x-1\right)^2}{\sqrt{x^2+x+1}+x-1}\lim\limits_{x\rightarrow+\infty}\frac{3x}{\sqrt{x^2+x+1}+x-1}=\lim\limits_{x\rightarrow+\infty}\frac{3}{\sqrt{1+\frac{1}{x}+\frac{1}{x^2}}+1-\frac{1}{x}}=\frac{3}{2}\)

\(L_8=\lim\limits_{x\rightarrow-\infty}\frac{x^2+x+1-\left(3x-2\right)^2}{\sqrt{x^2+x+1}+3x-2}=\lim\limits_{x\rightarrow-\infty}\frac{-8x^2+13x-3}{\sqrt{x^2+x+1}+3x-2}=\lim\limits_{x\rightarrow-\infty}\frac{-8+\frac{13}{x}-\frac{3}{x^2}}{-\sqrt{1+\frac{1}{x}+\frac{1}{x^2}}+3-\frac{2}{x}}=\frac{-8}{-1+3}=-4\)