Bài 1:

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

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

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

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

Bài 2:

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

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

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

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:

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

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:

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

\(\lim\limits_{x\to 1+}\frac{1}{x^2-1}.\lim\limits_{x\to 1+}(2x-1)=1.(+\infty)=+\infty \)

Tương tự \(\lim\limits_{x\to 1-} \frac{2x^2-3x+1}{x^3-x^2-x+1}=-\infty \)

a) <=> \(6x^2-5x+3-2x+3x\left(3-2x\right)=0\)

<=> \(6x^2-5x+3-2x+9x-6x^2=0\)

<=> \(2x+3=0\)

<=> \(x=\frac{-3}{2}\)

b) <=> \(10\left(x-4\right)-2\left(3+2x\right)=20x+4\left(1-x\right)\)

<=> \(10x-40-6-4x=20x+4-4x\)

<=> \(6x-46-16x-4=0\)

<=> \(-10x-50=0\)

<=> \(-10\left(x+5\right)=0\)

<=> \(x+5=0\)

<=> \(x=-5\)

c) <=> \(8x+3\left(3x-5\right)=18\left(2x-1\right)-14\)

<=> \(8x+9x-15=36x-18-14\)

<=> \(8x+9x-36x=+15-18-14\)

<=> \(-19x=-14\)

<=> \(x=\frac{14}{19}\)

d) <=>\(2\left(6x+5\right)-10x-3=8x+2\left(2x+1\right)\)

<=> \(12x+10-10x-3=8x+4x+2\)

<=> \(2x-7=12x+2\)

<=> \(2x-12x=7+2\)

<=> \(-10x=9\)

<=> \(x=\frac{-9}{10}\)

e) <=> \(x^2-16-6x+4=\left(x-4\right)^2\)

<=> \(x^2-6x-12-\left(x-4^2\right)=0\)

<=> \(x^2-6x-12-\left(x^2-8x+16\right)=0\)

<=> \(x^2-6x-12-x^2+8x-16=0\)

<=> \(2x-28=0\)

<=> \(2\left(x-14\right)=0\)

<=> x-14=0

<=> x=14

Luffy , cậu sai câu c nhé , kia là -17 ạ => x=17/19

1: \(\Leftrightarrow2x^2-10x-3x-2x^2=0\)

=>-13x=0

=>x=0

2: \(\Leftrightarrow5x-2x^2+2x^2-2x=13\)

=>3x=13

=>x=13/3

3: \(\Leftrightarrow4x^4-6x^3-4x^3+6x^3-2x^2=0\)

=>-2x^2=0

=>x=0

4: \(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)

=>-8x=6-14=-8

=>x=1

`1)2x(x-5)-(3x+2x^2)=0`

`<=>2x^2-10x-3x-2x^2=0`

`<=>-13x=0`

`<=>x=0`

___________________________________________________

`2)x(5-2x)+2x(x-1)=13`

`<=>5x-2x^2+2x^2-2x=13`

`<=>3x=13<=>x=13/3`

___________________________________________________

`3)2x^3(2x-3)-x^2(4x^2-6x+2)=0`

`<=>4x^4-6x^3-4x^4+6x^3-2x^2=0`

`<=>x=0`

___________________________________________________

`4)5x(x-1)-(x+2)(5x-7)=0`

`<=>5x^2-5x-5x^2+7x-10x+14=0`

`<=>-8x=-14`

`<=>x=7/4`

___________________________________________________

`5)6x^2-(2x-3)(3x+2)=1`

`<=>6x^2-6x^2-4x+9x+6=1`

`<=>5x=-5<=>x=-1`

___________________________________________________

`6)2x(1-x)+5=9-2x^2`

`<=>2x-2x^2+5=9-2x^2`

`<=>2x=4<=>x=2`

Bạn tự hiểu là giới hạn tiến đến đâu nhé, làm biếng gõ đủ công thức

a. \(\frac{\sqrt{1+x}-1+1-\sqrt[3]{1+x}}{x}=\frac{\frac{x}{\sqrt{1+x}+1}-\frac{x}{1+\sqrt[3]{1+x}+\sqrt[3]{\left(1+x\right)^2}}}{x}=\frac{1}{\sqrt{1+x}+1}-\frac{1}{1+\sqrt[3]{1+x}+\sqrt[3]{\left(1+x\right)^2}}=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}\)

b.

\(\frac{1-x^3-1+x}{\left(1-x\right)^2\left(1+x+x^2\right)}=\frac{x\left(1-x\right)\left(1+x\right)}{\left(1-x\right)^2\left(1+x+x^2\right)}=\frac{x\left(1+x\right)}{\left(1-x\right)\left(1+x+x^2\right)}=\frac{2}{0}=\infty\)

c.

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

d.

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

e.

\(=\frac{2x^2-8x+8}{\left(x-1\right)\left(x-2\right)\left(x-2\right)\left(x-3\right)}=\frac{2\left(x-2\right)^2}{\left(x-1\right)\left(x-3\right)\left(x-2\right)^2}=\frac{2}{\left(x-1\right)\left(x-3\right)}=\frac{2}{-1}=-2\)

f.

\(=\frac{2x}{x\sqrt{4+x}}=\frac{2}{\sqrt{4+x}}=1\)

1.

\(f\left(x\right)=\frac{x-7}{\left(x-4\right)\left(4x-3\right)}\)

Vậy:

\(f\left(x\right)\) ko xác định tại \(x=\left\{\frac{3}{4};4\right\}\)

\(f\left(x\right)=0\Rightarrow x=7\)

\(f\left(x\right)>0\Rightarrow\left[{}\begin{matrix}\frac{3}{4}< x< 4\\x>7\end{matrix}\right.\)

\(f\left(x\right)< 0\Rightarrow\left[{}\begin{matrix}x< \frac{3}{4}\\4< x< 7\end{matrix}\right.\)

2.

\(f\left(x\right)=\frac{11x+3}{-\left(x-\frac{5}{2}\right)^2-\frac{3}{4}}\)

Vậy:

\(f\left(x\right)=0\Rightarrow x=-\frac{3}{11}\)

\(f\left(x\right)>0\Rightarrow x< -\frac{3}{11}\)

\(f\left(x\right)< 0\Rightarrow x>-\frac{3}{11}\)

3.

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

Vậy:

\(f\left(x\right)\) ko xác định khi \(x=\left\{1;1\pm\sqrt{3}\right\}\)

\(f\left(x\right)=0\Rightarrow x=\frac{2}{3}\)

\(f\left(x\right)>0\Rightarrow\left[{}\begin{matrix}x< 1-\sqrt{3}\\\frac{2}{3}< x< 1\\x>1+\sqrt{3}\end{matrix}\right.\)

\(f\left(x\right)< 0\Rightarrow\left[{}\begin{matrix}1-\sqrt{3}< x< \frac{2}{3}\\1< x< 1+\sqrt{3}\end{matrix}\right.\)

4.

\(f\left(x\right)=\frac{\left(x-2\right)\left(x+6\right)}{\sqrt{6}\left(x+\frac{\sqrt{6}}{4}\right)^2+\frac{8\sqrt{2}-3\sqrt{6}}{8}}\)

Vậy:

\(f\left(x\right)=0\Rightarrow x=\left\{-6;2\right\}\)

\(f\left(x\right)>0\Rightarrow\left[{}\begin{matrix}x< -6\\x>2\end{matrix}\right.\)

\(f\left(x\right)< 0\Rightarrow-6< x< 2\)

5.

\(f\left(x\right)=\frac{x^2-3x-2}{-\left(x-\frac{1}{2}\right)^2-\frac{3}{4}}\)

Vậy:

\(f\left(x\right)=0\Rightarrow x=\frac{3\pm\sqrt{17}}{2}\)

\(f\left(x\right)>0\Rightarrow\frac{3-\sqrt{17}}{2}< x< \frac{3+\sqrt{17}}{2}\)

\(f\left(x\right)< 0\Rightarrow\left[{}\begin{matrix}x< \frac{3-\sqrt{17}}{2}\\x>\frac{3+\sqrt{17}}{2}\end{matrix}\right.\)

6.

\(f\left(x\right)=\frac{\left(x-1\right)\left(x^2+x-4\right)}{\left(x-1\right)^2\left(x^2-2x-5\right)}=\frac{x^2+x-4}{\left(x-1\right)\left(x^2-2x-5\right)}\)

Vậy:

\(f\left(x\right)\) ko xác định khi \(x=\left\{1;1\pm\sqrt{6}\right\}\)

\(f\left(x\right)=0\Rightarrow x=\left\{\frac{-1\pm\sqrt{17}}{2}\right\}\)

\(f\left(x\right)>0\Rightarrow\left[{}\begin{matrix}\frac{-1-\sqrt{17}}{2}< x< 1-\sqrt{6}\\1< x< \frac{-1+\sqrt{17}}{2}\\x>1+\sqrt{6}\end{matrix}\right.\)

\(f\left(x\right)< 0\Rightarrow\left[{}\begin{matrix}x< \frac{-1-\sqrt{17}}{2}\\1-\sqrt{6}< x< 1\\\frac{-1+\sqrt{17}}{2}< x< 1+\sqrt{6}\end{matrix}\right.\)

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

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

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

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

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

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

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

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

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

1/ \(=\lim\limits_{x\rightarrow0}\dfrac{3\left(1+3x\right)^2.3+4.4\left(1-4x\right)^3}{1}=...\left(thay-x-vo\right)\)

2/ \(=\lim\limits_{x\rightarrow2}\dfrac{2.2.x-5}{3x^2-3}=\dfrac{4.2-5}{3.4-3}=\dfrac{1}{3}\)

3/ \(=\lim\limits_{x\rightarrow1}\dfrac{4x^3-3}{3x^2+2}=\dfrac{4.1-3}{3.1-2}=1\)

Xai L'Hospital nhe :v

giup minh voi cac bạn