Ta có
\(lim_{x-->x0}=\frac{1}{\sqrt[2]{\left(0+1\right)^2+\sqrt[2]{0+1}+1}}=\frac{1}{\sqrt[2]{1^2+\sqrt[2]{1}+1}}=\frac{1}{\sqrt[2]{4}}=\frac{1}{2}\)
a/ \(^{lim}_{x->0}\frac{\sqrt{1+x}-\sqrt[3]{1+x}}{x}\)
b/\(^{lim}_{x->1}\left(\frac{1}{1-x}-\frac{1}{1-x^3}\right)\)
c/ \(^{lim}_{x->+\infty}\left(\sqrt[3]{2x-1}-\sqrt[3]{2x+1}\right)\)
d/ \(^{lim}_{x->-\infty}\left(\sqrt[3]{3x^3-1}+\sqrt{x^2+2}\right)\)
e/\(^{lim}_{x->2}\left(\frac{1}{x^2-3x+2}+\frac{1}{x^2-5x+6}\right)\)
f/ \(^{lim}_{x->0^{+-}}\left(\frac{2x}{\sqrt{4x^2+x^3}}\right)\)
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. \(\lim\limits_{x\rightarrow1}\frac{\sqrt[3]{3x-2}-\sqrt{2x-1}}{x^3-1}\)
d. \(\lim\limits_{x\rightarrow+\infty}\left(\sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x}\right)\) h. \(\lim\limits_{x\rightarrow1}\frac{\sqrt[3]{x+9}+\sqrt[3]{2x-6}}{x^3+1}\)
a,\(^{lim}_{x->2}\frac{\sqrt[3]{8x+11}-\sqrt{x+7}}{x^2-3x+2}\)
b, \(^{lim}_{x->0}\frac{2\sqrt{1+x}-\sqrt[3]{8-x}}{x}\)
c, \(^{lim}_{x->1}\frac{\sqrt{5-x^3}-\sqrt[3]{x^2+7}}{x^2-1}\)
d,\(^{lim}_{x->0}\frac{\sqrt{1+2x}.\sqrt[3]{1+4x}-1}{x}\)
e,\(^{lim}_{x->1}\frac{x^4-1}{x^3-2x^2+x}\)
f,\(^{lim}_{x->1}\left(\frac{1}{1-x}-\frac{3}{1-x^3}\right)\)
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
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. \(\lim\limits_{x\rightarrow2^+}\left(\frac{1}{x^2-3x+2}-\frac{1}{x^2-5x+6}\right)\)
Bài 1
a. \(\lim\limits_{x\rightarrow-\infty}\frac{\sqrt{4x^2}+1}{3x-1}\)
b. \(\lim\limits_{x\rightarrow+\infty}\frac{\sqrt{9x^2+x+1}-\sqrt{4x^2+2x+1}}{x+1}\)
c. \(\lim\limits_{x\rightarrow+\infty}\frac{\sqrt{x+2x+3}+4x+1}{\sqrt{4x^2+1}+2-x}\)
d. \(\lim\limits_{x\rightarrow+\infty}\frac{3x-2\sqrt{x}+\sqrt{x^4-5x}}{2x^2+4x-5}\)
Bài 2
a. \(\lim\limits_{x\rightarrow-\infty}\frac{2x+1}{x-1}\)
b. \(\lim\limits_{x\rightarrow-\infty}\frac{2x^3+3}{x^3-2x^2+1}\)
c. \(\lim\limits_{x\rightarrow+\infty}\frac{\left(3x^2+1\right)\left(5x+3\right)}{\left(2x^3-1\right)\left(x+4\right)}\)
Tính các giới hạn sau đây :
\(L_1=lim\frac{x^3+3x^2-2x}{x^5+4x}\left(x\rightarrow0\right)\)
\(L_2=lim\frac{x^3-3x+2}{\left(4-2x\right)^3}\left(x\rightarrow+\infty\right)\)
\(L_3=lim\frac{2x^2+3x+1}{x^2+x}\left(x\rightarrow-1\right)\)
\(L_4=lim\frac{x^2-4x+1}{4-x^2}\left(x\rightarrow2\right)\)
\(L_5=lim\frac{\sqrt{x+1}-2}{x-2}\left(x\rightarrow3\right)\)
\(L_6=lim\frac{\sqrt{x+3}-x-1}{x^2-1}\left(x\rightarrow1\right)\)
\(L_7=lim\left(\sqrt{x^2+x+1}-x+1\right)\left(x\rightarrow+\infty\right)\)
\(L_8=lim\left(\sqrt{x^2+x+1}-3x+2\right)\left(x\rightarrow-\infty\right)\)
Tính các giới hạn sau:\(I_1=\lim\limits_{x\rightarrow1}\dfrac{\left(1-\sqrt{x}\right)\left(1-\sqrt[3]{x}\right)....\left(1-\sqrt[n]{x}\right)}{\left(1-x\right)^{n-1}}\)
\(I_2=\lim\limits_{x\rightarrow0}\dfrac{\left(\sqrt{1+x^2}+x\right)^n-\left(\sqrt{1+x^2}-x\right)^n}{x}\)
Giá trị của các giới hạn :
a, lim\(\left(\sqrt[3]{3x^3-1}+\sqrt{x^2+1}\right)\) khi x→\(-\infty\)
b, lim\(\left(\sqrt{x^2+x}-\sqrt[3]{x^3-x^2}\right)\) khi x→\(+\infty\)
c, lim\(\left(\sqrt[3]{2x-1}-\sqrt[3]{2x+1}\right)\) khi x→\(+\infty\)
