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Trần Trọng Thái
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Hoàng Tử Hà
18 tháng 2 2021 lúc 1:37

a/ \(=\lim\limits_{x\rightarrow-\infty}\dfrac{x^2+1-x^2}{\sqrt{x^2+1}-x}+\lim\limits_{x\rightarrow-\infty}\dfrac{3x^3-1-x^3}{\sqrt[3]{\left(3x^3-1\right)^2}+x\sqrt[3]{3x^3-1}+x^2}\)

\(=\lim\limits_{x\rightarrow-\infty}\dfrac{\dfrac{1}{x}}{-\sqrt{\dfrac{x^2}{x^2}+\dfrac{1}{x^2}}-\dfrac{x}{x}}+\lim\limits_{x\rightarrow-\infty}\dfrac{-\dfrac{1}{x^2}}{\dfrac{\sqrt[3]{\left(3x^3-1\right)^2}}{x^2}+\dfrac{x\sqrt[3]{3x^3-1}}{x^2}+\dfrac{x^2}{x^2}}=0\)

b/ \(=\lim\limits_{x\rightarrow+\infty}\dfrac{x^2+x-x^2}{\sqrt{x^2+x}+x}+\lim\limits_{x\rightarrow+\infty}\dfrac{x^3-x^3+x^2}{x^2+x\sqrt[3]{x^3-x^2}+\sqrt[3]{\left(x^3-x^2\right)^2}}\)

\(=\lim\limits_{x\rightarrow+\infty}\dfrac{\dfrac{x}{x}}{\sqrt{\dfrac{x^2}{x^2}+\dfrac{x}{x^2}}+\dfrac{x}{x}}+\lim\limits_{x\rightarrow+\infty}\dfrac{\dfrac{x^2}{x^2}}{\dfrac{x^2}{x^2}+\dfrac{x\sqrt[3]{x^3-x^2}}{x^2}+\dfrac{\sqrt[3]{\left(x^3-x^2\right)^2}}{x^2}}\)

\(=\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}\)

c/ \(=\lim\limits_{x\rightarrow+\infty}\dfrac{2x-1-2x-1}{\sqrt[3]{\left(2x-1\right)^2}+\sqrt[3]{4x^2-1}+\sqrt[3]{\left(2x+1\right)^2}}\)

\(=\lim\limits_{x\rightarrow+\infty}\dfrac{-\dfrac{2}{x^{\dfrac{2}{3}}}}{\dfrac{\sqrt[3]{\left(2x-1\right)^2}}{x^{\dfrac{2}{3}}}+\dfrac{\sqrt[3]{4x^2-1}}{x^{\dfrac{2}{3}}}+\dfrac{\sqrt[3]{\left(2x+1\right)^2}}{x^{\dfrac{2}{3}}}}=0\)

Check lai ho minh nhe :v

sgfr hod
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Nguyễn Lê Phước Thịnh
6 tháng 12 2023 lúc 21:59

\(\lim\limits_{x\rightarrow-\infty}f\left(x\right)=\lim\limits_{x\rightarrow-\infty}\dfrac{\sqrt{x^2-3}}{x+3}\)

\(=\lim\limits_{x\rightarrow-\infty}\dfrac{\sqrt{x^2\left(1-\dfrac{3}{x^2}\right)}}{x+3}\)

\(=\lim\limits_{x\rightarrow-\infty}\dfrac{-x\cdot\sqrt{1-\dfrac{3}{x^2}}}{x\left(1+\dfrac{3}{x}\right)}=\lim\limits_{x\rightarrow-\infty}\dfrac{-\sqrt{1-\dfrac{3}{x^2}}}{1+\dfrac{3}{x}}\)

\(=\dfrac{-\sqrt{1-0}}{1+0}=-\dfrac{1}{1}=-1\)

Julian Edward
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Nguyễn Việt Lâm
2 tháng 3 2021 lúc 22:32

\(\lim\limits_{x\rightarrow-\infty}\dfrac{-\sqrt{1-\dfrac{3}{x}+\dfrac{6}{x^2}}+2}{2-\dfrac{3}{x}}=\dfrac{-1+2}{2}=\dfrac{1}{2}\)

Julian Edward
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Nguyễn Việt Lâm
8 tháng 3 2021 lúc 23:40

\(\lim\limits_{x\rightarrow1^+}f\left(x\right)=\lim\limits_{x\rightarrow1^-}f\left(x\right)\Rightarrow\lim\limits_{x\rightarrow1}f\left(x\right)=2\)

Julian Edward
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Nguyễn Việt Lâm
2 tháng 3 2021 lúc 22:33

\(=\lim\limits_{x\rightarrow-\infty}\dfrac{1-\sqrt{4x^2-x+5}}{-ax+2}=\lim\limits_{x\rightarrow-\infty}\dfrac{\dfrac{1}{x}+\sqrt{4-\dfrac{1}{x}+\dfrac{5}{x^2}}}{-a+\dfrac{2}{x}}=\dfrac{2}{-a}=\dfrac{2}{3}\)

\(\Rightarrow a=-3\)

Julian Edward
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Nguyễn Việt Lâm
27 tháng 1 2021 lúc 19:00

\(\lim\limits_{x\rightarrow-\infty}\dfrac{\left(2x^2-1\right)\left(mx+3\right)}{x^3+4x+7}=\lim\limits_{x\rightarrow-\infty}\dfrac{\left(2-\dfrac{1}{x^2}\right)\left(m+\dfrac{3}{x}\right)}{1+\dfrac{4}{x^2}+\dfrac{7}{x^3}}=2m\)

\(\Rightarrow2m=6\Rightarrow m=3\)

Tâm Cao
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Hoàng Tử Hà
4 tháng 4 2021 lúc 0:31

Xet \(m\ne-3\)

\(=\lim\limits_{x\rightarrow-\infty}x\left(\sqrt[3]{1}+\sqrt{4}+m\right)=x\left(3+m\right)\)

\(=\left[{}\begin{matrix}-\infty\left(m>-3\right)\\+\infty\left(m< -3\right)\end{matrix}\right.\)

Xet \(m=-3\)

\(=\lim\limits_{x\rightarrow-\infty}\left(\sqrt[3]{x^3+2x^2+1}-x-2x-\sqrt{4x^2+2x+3}\right)\)

\(=\lim\limits_{x\rightarrow-\infty}\dfrac{x^3+2x^2+1-x^3}{\sqrt[3]{\left(x^3+2x^2+1\right)^2}+x\sqrt[3]{x^3+2x^2+1}+x^2}-\lim\limits_{x\rightarrow-\infty}\dfrac{4x^2-4x^2-2x-3}{2x-\sqrt{4x^2+2x+3}}\)

\(=\dfrac{2}{3}+\dfrac{1}{2}=\dfrac{7}{6}\)

Châu Ngọc Minh Anh
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Hoàng Tử Hà
19 tháng 2 2021 lúc 11:56

\(\lim\limits_{x\rightarrow-\infty}\dfrac{x^2+ax+5-x^2}{\sqrt{x^2+ax+5}-x}=\lim\limits_{x\rightarrow-\infty}\dfrac{\dfrac{ax}{x}+\dfrac{5}{x}}{-\sqrt{\dfrac{x^2}{x^2}+\dfrac{ax}{x^2}+\dfrac{5}{x^2}}-\dfrac{x}{x}}=\dfrac{-a}{2}\)

\(-\dfrac{a}{2}=5\Rightarrow a=-10\)

sgfr hod
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Nguyễn Lê Phước Thịnh
6 tháng 12 2023 lúc 20:00

\(\lim\limits_{x\rightarrow-\infty}f\left(x\right)\)

=\(\lim\limits_{x\rightarrow-\infty}\dfrac{2x-1}{\sqrt{x^2+1}-1}\)

\(=\lim\limits_{x\rightarrow-\infty}\dfrac{x\left(2-\dfrac{1}{x}\right)}{-x\cdot\sqrt{1+\dfrac{1}{x^2}}-1}\)

\(=\lim\limits_{x\rightarrow-\infty}\dfrac{2-\dfrac{1}{x}}{-\sqrt{1+\dfrac{1}{x^2}}-\dfrac{1}{x}}=\dfrac{2-0}{-\sqrt{1+0}-0}=\dfrac{2}{-1}=-2\)

Hoàng minh
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