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 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 \(\lim\limits_{x\to 1-} \frac{2x^2-3x+1}{x^3-x^2-x+1}=-\infty \)
Bài 10:
\(\lim\limits_{x\to -\infty}\frac{(2x^2+1)^2(5x+3)}{(2x^3-1)(x+1)^2}=\lim\limits_{x\to -\infty}\frac{(\frac{2x^2+1}{x^2})^2.\frac{5x+3}{x}}{\frac{2x^3-1}{x^3}.(\frac{x+1}{x})^2}=\lim\limits_{x\to -\infty}\frac{(2+\frac{1}{x^2})^2(5+\frac{3}{x})}{(2-\frac{1}{x^3})(1+\frac{1}{x})^2}=\frac{2^2.5}{2.1}=10\)
Nốt nhé hôm qua mình buồn ngủ nên không làm tiếp được
Bài 11:
\(\lim\limits_{x\to -\infty}\frac{\sqrt{x^2+2x}}{x+3}=\lim\limits_{x\to -\infty}\frac{\frac{\sqrt{x^2+2x}}{-x}}{\frac{x+3}{-x}}\)\(=\lim\limits_{x\to -\infty}\frac{\sqrt{1+\frac{2}{x}}}{-1+\frac{-3}{x}}=\frac{1}{-1}=-1\)
Bài 12:
\(\lim\limits_{x\to 1}\frac{\sqrt{5-x^3}-\sqrt[3]{x^2+7}}{x^2-1}=\lim\limits_{x\to 1}\frac{(\sqrt{5-x^3}-2)-(\sqrt[3]{x^2+7}-2)}{(x-1)(x+1)}\)
\(=\lim\limits_{x\to 1}\frac{\frac{1-x^3}{\sqrt{5-x^3}+2}-\frac{x^2-1}{\sqrt[3]{(x^2+7)^2}+2\sqrt[3]{x^2+7}+4}}{(x-1)(x+1)}\)
\(=\lim\limits_{x\to 1}\frac{\frac{-(x^2+x+1)}{\sqrt{5-x^3}+2}-\frac{x+1}{\sqrt[3]{(x^2+7)^2}+2\sqrt[3]{x^2+7}+4}}{x+1}=\frac{-11}{24}\)
Bài 13:
\(\lim\limits _{x\to 0}\frac{\sqrt[3]{x+1}+\sqrt{x+4}-3}{x}=\lim\limits _{x\to 0}\frac{(\sqrt[3]{x+1}-1)+(\sqrt{x+4}-2)}{x}=\lim\limits _{x\to 0}\frac{\frac{x}{\sqrt[3]{(x+1)^2}+\sqrt[3]{x+1}+1}+\frac{x}{\sqrt{x+4}+2}}{x}\)
\(\lim\limits _{x\to 0}[\frac{1}{\sqrt[3]{(x+1)^2}+\sqrt[3]{x+1}+1}+\frac{1}{\sqrt{x+4}+2}]=\frac{7}{12}\)
Bài 14:
\(\lim\limits_{x\to 0}\frac{(x^2+2020)\sqrt{1+3x}-2020}{x}=\lim\limits_{x\to 0}\frac{x^2\sqrt{1+3x}+2020(\sqrt{1+3x}-1)}{x}\)
\(\lim\limits_{x\to 0}x\sqrt{1+3x}+2020\lim\limits_{x\to 0}\frac{3x}{(\sqrt{1+3x}+1)x}=\lim\limits_{x\to 0}x\sqrt{1+3x}+2020\lim\limits_{x\to 0}\frac{3}{\sqrt{1+3x}+1}\)
\(=0+2020.\frac{3}{2}=3030\)
Bài 15:
\(\lim\limits_{x\to +\infty}(2x-\sqrt{4x^2-3})=\lim\limits_{x\to +\infty}\frac{4x^2-(4x^2-3)}{2x+\sqrt{4x^2-3}}=\lim\limits_{x\to +\infty}\frac{3}{2x+\sqrt{4x^2-3}}=0\)
Bài 16:
\(\lim\limits _{x\to a}\frac{x^2-(a+1)x+a}{x^3-a^3}=\lim\limits _{x\to a}\frac{(x-1)(x-a)}{(x^2+xa+a^2)(x-a)}=\lim\limits _{x\to a}\frac{x-1}{x^2+xa+a^2}\)
Nếu $a=0$ thì \(=\lim\limits _{x\to 0}\frac{x-1}{x^2}=\lim\limits _{x\to 0}\frac{1}{x^2}.\lim\limits _{x\to 0}(x-1)=+\infty (-1)=-\infty \)
Nếu $a\neq 0$ thì \(\lim\limits _{x\to a}\frac{x-1}{x^2+xa+a^2}=\frac{a-1}{3a^2}\)
Bài 17: Áp dụng công thức L'Hospital
\(\lim\limits _{x\to 1}\frac{x^n-nx+n-1}{(x-1)^2}=\lim\limits _{x\to 1}\frac{dx(x^n-nx+n-1)}{dx((x-1)^2}=\lim\limits _{x\to 1}\frac{nx^{n-1}-n}{2x-2}\)
\(=\lim\limits _{x\to 1}\frac{dx(nx^{n-1}-n)}{dx(2x-2)}=\lim\limits _{x\to 1}\frac{n(n-1)x^{n-2}}{2}=\frac{n(n-1)}{2}\)
Bài 18:
Bài 18:
\(\lim\limits _{x\to 2+}f(x)=\lim\limits _{x\to 2+}\frac{x^2-2x}{8-x^3}=\lim\limits _{x\to 2+}\frac{x(x-2)}{(2-x)(4+2x+x^2)}=\lim\limits _{x\to 2+}\frac{-x}{x^2+2x+4}=\frac{-1}{6}\)
\(\lim\limits _{x\to 2-}f(x)=\lim\limits _{x\to 2-}\frac{x^4-16}{x-2}=\lim\limits _{x\to 2-}\frac{(x^2-4)(x^2+4)}{x-2}=\lim\limits _{x\to 2-}(x+2)(x^2+4)=32\)
Ta thấy \(\lim\limits _{x\to 2+}f(x)\neq \lim\limits _{x\to 2-}f(x)\) nên không tồn tại giới hạn $f(x)$ tại $x=2$
Bài 1:
\(\lim\limits_{x\to1}\frac{2x^2-3x+1}{x^3-x^2-x+1}=\lim\limits_{x\to1}\frac{\left(x-1\right)\left(2x-1\right)}{\left(x+1\right)\left(x-1\right)^2}=\lim\limits_{x\to1}\frac{2x-1}{\left(x+1\right)\left(x-1\right)}\)
\(=\lim\limits_{x\to 1}\frac{2(x-1)+1}{(x-1)(x+1)}=\lim\limits_{x\to 1}\frac{2}{x+1}+\lim\limits_{x\to 1}\frac{1}{x^2-1}\)
Có: \(\lim\limits_{x\to 1+} \frac{2x^2-3x+1}{x^3-x^2-x+1}=+\infty \)
\(\lim\limits_{x\to 1-} \frac{2x^2-3x+1}{x^3-x^2-x+1}=-\infty \)
