Question

Tính các giới hạn sau: (2 điểm)

a. $\underset{x\to 1}{\mathop{\lim }}\,\frac{{{x}^{3}}+2x-3}{{{x}^{2}}-x}$;

b. $\underset{x\to 1}{\mathop{\lim }}\,\frac{\sqrt{2x+2}-\sqrt{3x+1}}{x-1}$.

Answer 1

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

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

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

\(=\frac{1^2+1+3}{1}\)   

\(=5\)   

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

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

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

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

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

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

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

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

Answer 2

  

Answer 3

https://drive.google.com/file/d/14Q-YI3szy-rePnIHWGD35RKCWiCXCT6k/view?usp=sharing

https://drive.google.com/file/d/1425SNt8hu4qt2y1kIcnhIvcxPfODsY1T/view?usp=sharing

Answer 4

  

Answer 5

Võ Ngọc Tú Uyên-41    

Answer 6

Answer 7

a. lim_x→1 = 5

b. lim_x→1 = -1/4

Answer 8

  Nguyễn Minh Anh   

Answer 9

Lê Thị Anh Thư  

Answer 10

  

Answer 11

Dương Thùy Dung  

Answer 12

  

Answer 13