b)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời. 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: \(a=\lim\limits_{x\rightarrow-\infty}\frac{2+\frac{1}{x}}{1-\frac{1}{x}}=2\) \(b=\lim\limits_{x\rightarrow-\infty}\frac{2+\frac{3}{x^3}}{1-\frac{2}{x}+\frac{1}{x^3}}=2\) \(c=\lim\limits_{x\rightarrow+\infty}\frac{x^2\left(3+\frac{1}{x^2}\right)x\left(5+\frac{3}{x}\right)}{x^3\left(2-\frac{1}{x^3}\right)x\left(1+\frac{4}{x}\right)}=\frac{15}{+\infty}=0\) \(a=\lim\limits_{x\rightarrow2}\frac{\sqrt[3]{8x+11}-3+3-\sqrt{x+7}}{\left(x-1\right)\left(x-2\right)}=\lim\limits_{x\rightarrow2}\frac{\frac{8\left(x-2\right)}{\sqrt[3]{\left(8x+11\right)^2}+3\sqrt[3]{8x+11}+9}-\frac{x-2}{9+\sqrt{x+7}}}{\left(x-1\right)\left(x-2\right)}\) \(=\lim\limits_{x\rightarrow2}\frac{\frac{8}{\sqrt[3]{\left(8x+11\right)^2}+3\sqrt[3]{8x+11}+9}-\frac{1}{9+\sqrt{x+7}}}{x-1}=\frac{29}{36}\) \(b=\lim\limits_{x\rightarrow+\infty}\frac{x^2\left(2-\frac{3}{x}\right)^2.x^3\left(4+\frac{7}{x}\right)^3}{x^3\left(3+\frac{1}{x^3}\right).x^2\left(10+\frac{9}{x^2}\right)}=\frac{2.4}{3.10}=\frac{4}{15}\) \(c=\lim\limits_{x\rightarrow0}\frac{\sqrt{1+4x}-\left(2x+1\right)+\left(2x+1-\sqrt[3]{1+6x}\right)}{x^2}\) \(=\lim\limits_{x\rightarrow0}\frac{\frac{-4x^2}{\sqrt{1+4x}+2x+1}+\frac{8x^3+12x^2}{\left(2x+1\right)^2+\left(2x+1\right)\sqrt[3]{1+6x}+\sqrt[3]{\left(1+6x\right)^2}}}{x^2}\) \(=\lim\limits_{x\rightarrow0}\left(\frac{-4}{\sqrt{1+4x}+2x+1}+\frac{8x+12}{\left(2x+1\right)^2+\left(2x+1\right)\sqrt[3]{1+6x}+\sqrt[3]{\left(1+6x\right)^2}}\right)=\frac{-4}{1+1}+\frac{12}{1+1+1}=2\) \(d=\lim\limits_{x\rightarrow0}\frac{\sqrt{1+6x}\left(\sqrt{1+4x}-1\right)}{x}+\lim\limits_{x\rightarrow0}\frac{\sqrt{1+6x}-1}{x}\) \(=\lim\limits_{x\rightarrow0}\frac{4x\sqrt{1+6x}}{x\left(\sqrt{1+4x}+1\right)}+\lim\limits_{x\rightarrow0}\frac{6x}{x\left(\sqrt{1+6x}+1\right)}\) \(=\lim\limits_{x\rightarrow0}\frac{4\sqrt{1+6x}}{\sqrt{1+4x}+1}+\lim\limits_{x\rightarrow0}\frac{6}{\sqrt{1+6x}+1}=\frac{4}{1+1}+\frac{6}{1+1}=5\) \(e=\lim\limits_{x\rightarrow0}\frac{\sqrt[3]{1+4x}\left(\sqrt{1+2x}-1\right)}{x}+\lim\limits_{x\rightarrow0}\frac{\sqrt[3]{1+4x}-1}{x}\) \(=\lim\limits_{x\rightarrow0}\frac{2x\sqrt[3]{1+4x}}{x\left(\sqrt{1+2x}+1\right)}+\lim\limits_{x\rightarrow0}\frac{4x}{x\left(\sqrt[3]{\left(1+4x\right)^2}+\sqrt[3]{1+4x}+1\right)}\) \(=\lim\limits_{x\rightarrow0}\frac{2\sqrt[3]{1+4x}}{\sqrt{1+2x}+1}+\lim\limits_{x\rightarrow0}\frac{4}{\sqrt[3]{\left(1+4x\right)^2}+\sqrt[3]{1+4x}+1}=\frac{2}{1+1}+\frac{4}{1+1+1}=\frac{7}{3}\) 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. \(=\frac{2x}{x\sqrt{4+x}}=\frac{2}{\sqrt{4+x}}=1\) 16. \(y'=\frac{\left(cos2x\right)'}{2\sqrt{cos2x}}=\frac{-2sin2x}{2\sqrt{cos2x}}=-\frac{sin2x}{\sqrt{cos2x}}\) 17. \(y'=4x^3-\frac{1}{x^2}-\frac{1}{2\sqrt{x}}\) 18. \(y'=3x^2-2x\) \(y'\left(-2\right)=16;y\left(-2\right)=-12\) Pttt: \(y=16\left(x+2\right)-12\Leftrightarrow y=16x+20\) 19. \(y'=-\frac{1}{x^2}=-x^{-2}\) \(y''=2x^{-3}=\frac{2}{x^3}\) 20. \(\left(cotx\right)'=-\frac{1}{sin^2x}\) 21. \(y'=1+\frac{4}{x^2}=\frac{x^2+4}{x^2}\) 22. \(lim\left(3^n\right)=+\infty\) 11. \(\lim\limits_{x\rightarrow1^+}\frac{-2x+1}{x-1}=\frac{-1}{0}=-\infty\) 12. \(y=cotx\Rightarrow y'=-\frac{1}{sin^2x}\) 13. \(y'=2020\left(x^3-2x^2\right)^{2019}.\left(x^3-2x^2\right)'=2020\left(x^3-2x^2\right)^{2019}\left(3x^2-4x\right)\) 14. \(y'=\frac{\left(4x^2+3x+1\right)'}{2\sqrt{4x^2+3x+1}}=\frac{8x+3}{2\sqrt{4x^2+3x+1}}\) 15. \(y'=4\left(x-5\right)^3\) a) Cách 1: y' = (9 -2x)'(2x3- 9x2 +1) +(9 -2x)(2x3- 9x2 +1)' = -2(2x3- 9x2 +1) +(9 -2x)(6x2 -18x) = -16x3 +108x2 -162x -2. Cách 2: y = -4x4 +36x3 -81x2 -2x +9, do đó y' = -16x3 +108x2 -162x -2. b) y' = c) y' = (x -2)'√(x2 +1) + (x -2)(√x2 +1)' = √(x2 +1) + (x -2) d) y' = 2tanx.(tanx)' - (x2)' e) y' = \(L_1=\lim\limits_{x\rightarrow0}\frac{x\left(x^2+3x-2\right)}{x\left(x^4+4\right)}=\lim\limits_{x\rightarrow0}\frac{x^2+3x-2}{x^4+4}=-\frac{1}{2}\) \(L_2=\lim\limits_{x\rightarrow+\infty}\frac{1-\frac{3}{x^2}+\frac{2}{x^3}}{\left(\frac{4}{x}-2\right)^3}=\frac{1}{\left(-2\right)^3}=-\frac{1}{8}\) \(L_3=\lim\limits_{x\rightarrow-1}\frac{\left(2x+1\right)\left(x+1\right)}{x\left(x+1\right)}=\lim\limits_{x\rightarrow-1}\frac{2x+1}{x}=1\) \(L_4=\lim\limits_{x\rightarrow2}\frac{x^2-4x+1}{4-x^2}=\frac{1}{0}=+\infty\) \(L_5=\lim\limits_{x\rightarrow3}\frac{\sqrt{x+1}-2}{x-2}=\frac{0}{1}=0\) \(L_6=\lim\limits_{x\rightarrow1}\frac{x+3-\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)\left(\sqrt{x+3}+x+1\right)}=\lim\limits_{x\rightarrow1}\frac{-\left(x-1\right)\left(x+2\right)}{\left(x-1\right)\left(x+1\right)\left(\sqrt{x+3}+x+1\right)}\) \(=\lim\limits_{x\rightarrow1}\frac{-\left(x+2\right)}{\left(x+1\right)\left(\sqrt{x+3}+x+1\right)}=\frac{-3}{2.4}=-\frac{3}{8}\) \(L_7=\lim\limits_{x\rightarrow+\infty}\frac{x^2+x+1-\left(x-1\right)^2}{\sqrt{x^2+x+1}+x-1}\lim\limits_{x\rightarrow+\infty}\frac{3x}{\sqrt{x^2+x+1}+x-1}=\lim\limits_{x\rightarrow+\infty}\frac{3}{\sqrt{1+\frac{1}{x}+\frac{1}{x^2}}+1-\frac{1}{x}}=\frac{3}{2}\) \(L_8=\lim\limits_{x\rightarrow-\infty}\frac{x^2+x+1-\left(3x-2\right)^2}{\sqrt{x^2+x+1}+3x-2}=\lim\limits_{x\rightarrow-\infty}\frac{-8x^2+13x-3}{\sqrt{x^2+x+1}+3x-2}=\lim\limits_{x\rightarrow-\infty}\frac{-8+\frac{13}{x}-\frac{3}{x^2}}{-\sqrt{1+\frac{1}{x}+\frac{1}{x^2}}+3-\frac{2}{x}}=\frac{-8}{-1+3}=-4\) 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: \(\lim\limits _{x\to -1}\frac{\sqrt[3]{x}+1}{\sqrt{x^2+3}-2}=\lim\limits _{x\to -1}\frac{x+1}{\sqrt[3]{x^2}-\sqrt[3]{x}+1}.\frac{\sqrt{x^2+3}+2}{x^2+3-4}=\lim\limits _{x\to -1}\frac{x+1}{\sqrt[3]{x^2}-\sqrt[3]{x}+1}.\frac{\sqrt{x^2+3}+2}{(x-1)(x+1)}\) \(\lim\limits _{x\to -1}\frac{\sqrt{x^2+3}+2}{(\sqrt[3]{x^2}-\sqrt[3]{x}+1)(x-1)}=\frac{-2}{3}\)
K
Khách
NL
Nguyễn Lê Phước Thịnh
13 tháng 11 2023
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