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,...
Đọc tiếp
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, \(\lim\limits_{x\rightarrow-\infty}\left(8+4x-x^3\right)\)
9, \(\lim\limits_{x\rightarrow-1}\frac{\sqrt[3]{x}+1}{\sqrt{x^2+3}-2}\)
10, \(\lim\limits_{x\rightarrow-\infty}\frac{\left(2x^2+1\right)^2\left(5x+3\right)}{\left(2x^3-1\right)\left(x+1\right)^2}\)
11, \(\lim\limits_{x\rightarrow-\infty}\frac{\sqrt{x^2+2x}}{x+3}\)
12, \(\lim\limits_{x\rightarrow1}\frac{\sqrt{5-x^3}-\sqrt[3]{x^2+7}}{x^2-1}\)
13, \(\lim\limits_{x\rightarrow0}\frac{\sqrt[3]{x+1}+\sqrt{x+4}-3}{x}\)
14, \(\lim\limits_{x\rightarrow0}\frac{\left(x^2+2020\right)\sqrt{1+3x}-2020}{x}\)
15, \(\lim\limits_{x\rightarrow+\infty}\left(2x-\sqrt{4x^2-3}\right)\)
16, \(\lim\limits_{x\rightarrow a}\frac{x^2-\left(a+1\right)x+a}{x^3-a^3}\)
17, \(\lim\limits_{x\rightarrow1}\frac{x^n-nx+n-1}{\left(x-1\right)^2}\)
18, \(f\left(x\right)=\left\{{}\begin{matrix}\frac{x^2-2x}{8-x^3}\\\frac{x^4-16}{x-2}\end{matrix}\right.\) khi x>2,khi x<2 tại 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.
\(=\frac{2x}{x\sqrt{4+x}}=\frac{2}{\sqrt{4+x}}=1\)
cậu giúp mình bài mình mới đăng đc ko ạ