\(\infty\)x\(\infty\dfrac{1}{3}\)
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HT
27 tháng 1 2021
a/ \(=\lim\limits_{x\rightarrow-\infty}\dfrac{-x\sqrt{\dfrac{4x^2}{x^2}-\dfrac{2}{x^2}}-x\sqrt[3]{\dfrac{x^3}{x^3}+\dfrac{1}{x^3}}}{-x\sqrt{\dfrac{x^2}{x^2}+\dfrac{1}{x^2}}-x}\)
\(=\lim\limits_{x\rightarrow-\infty}\dfrac{-\sqrt{4}-1}{-1-1}=\dfrac{3}{2}\)
b/ \(=\lim\limits_{x\rightarrow-\infty}\dfrac{\dfrac{2x}{x}+\dfrac{3}{x}}{-\sqrt{\dfrac{2x^2}{x^2}-\dfrac{3}{x^2}}}=\dfrac{2}{-\sqrt{2}}=-\sqrt{2}\)
c/ \(\lim\limits_{x\rightarrow\pm\infty}\dfrac{\dfrac{2x^2}{x^2}-\dfrac{1}{x^2}}{\dfrac{3}{x^2}-\dfrac{x^2}{x^2}}=\dfrac{2}{-1}=-2\)
NV
Nguyễn Việt Lâm
Giáo viên
26 tháng 3 2022
\(\lim\limits_{x\rightarrow+\infty}\dfrac{17}{x^2+1}=\lim\limits_{x\rightarrow+\infty}\dfrac{\dfrac{17}{x^2}}{1+\dfrac{1}{x^2}}=\dfrac{0}{1}=0\)
\(\lim\limits_{x\rightarrow+\infty}\dfrac{-2x^2+x-1}{3+x}=\lim\limits_{x\rightarrow+\infty}x\left(\dfrac{-2+\dfrac{1}{x}-\dfrac{1}{x^2}}{\dfrac{3}{x}+1}\right)\)
Do \(\lim\limits_{x\rightarrow+\infty}x=+\infty\)
\(\lim\limits_{x\rightarrow+\infty}\left(\dfrac{-2+\dfrac{1}{x}-\dfrac{1}{x^2}}{\dfrac{3}{x}+1}\right)=-2< 0\)
\(\Rightarrow\lim\limits_{x\rightarrow+\infty}x\left(\dfrac{-2+\dfrac{1}{x}-\dfrac{1}{x^2}}{\dfrac{3}{x}+1}\right)=-\infty\)
HT
9 tháng 2 2021
1/ \(=\lim\limits_{x\rightarrow+\infty}\dfrac{\dfrac{2x}{x}-\sqrt{\dfrac{3x^2}{x^2}+\dfrac{2}{x^2}}}{\dfrac{5x}{x}+\sqrt{\dfrac{x^2}{x^2}+\dfrac{1}{x^2}}}=\dfrac{2-\sqrt{3}}{5+1}=\dfrac{2-\sqrt{3}}{6}\)
2/ \(=\lim\limits_{x\rightarrow+\infty}\sqrt{\dfrac{\dfrac{x^2}{x^4}+\dfrac{1}{x^4}}{\dfrac{2x^4}{x^4}+\dfrac{x^2}{x^4}-\dfrac{3}{x^4}}}=0\)
3/ \(=\lim\limits_{x\rightarrow-\infty}\dfrac{-\sqrt[3]{\dfrac{x^6}{x^6}+\dfrac{x^4}{x^6}+\dfrac{1}{x^6}}}{\sqrt{\dfrac{x^4}{x^4}+\dfrac{x^3}{x^4}+\dfrac{1}{x^4}}}=-1\)
HT
9 tháng 2 2021
Da nan roi mang meo lam mat het bai -.-
1/ \(=\lim\limits_{x\rightarrow-\infty}\dfrac{\sqrt[3]{\dfrac{3x^3}{x^3}+\dfrac{1}{x^3}}+\sqrt{\dfrac{2x^2}{x^2}+\dfrac{x}{x^2}+\dfrac{1}{x^2}}}{-\sqrt[4]{\dfrac{4x^4}{x^4}+\dfrac{2}{x^4}}}=\dfrac{-\sqrt[3]{3}-\sqrt{2}}{\sqrt[4]{4}}\)
2/ \(=\lim\limits_{x\rightarrow+\infty}\dfrac{8x^7}{\left(-2x^7\right)}=-\dfrac{8}{2^7}\)
3/ \(=\lim\limits_{x\rightarrow+\infty}\dfrac{\left(4x^2-3x+4-4x^2\right)\left(\sqrt{x^2+x+1}+x\right)}{\left(x^2+x+1-x^2\right)\left(\sqrt{4x^2-3x+4}+2x\right)}=\dfrac{-3.2}{2}=-3\)
2
18 tháng 11 2023
`a)lim_{x->+oo}[x+1]/[x^2+x+1]`
`=lim_{x->+oo}[1/x+1/[x^2]]/[1+1/x+1/[x^2]]`
`=0`
`b)lim_{x->+oo}[3x+1]/[3x^2-x+5]`
`=lim_{x->+oo}[3/x+1/[x^2]]/[3-1/x+5/[x^2]]`
`=0`
`c)lim_{x->-oo}[3x+5]/[\sqrt{x^2+x}]`
`=lim_{x->-oo}[3+5/x]/[-\sqrt{1+1/x}]`
`=-3`
`d)lim_{x->+oo}[-5x+1]/[\sqrt{3x^2+1}]`
`=lim_{x->+oo}[-5+1/x]/[\sqrt{3+1/[x^2]}]`
`=-5/3`
NV
Nguyễn Việt Lâm
Giáo viên
1 tháng 8 2021
\(y=x+\dfrac{1}{x}-5\ge2\sqrt{\dfrac{x}{x}}-5=-3\)
\(y_{min}=-3\) khi \(x=1\)
\(y=4x^2+\dfrac{1}{2x}+\dfrac{1}{2x}-4\ge3\sqrt[3]{\dfrac{4x^2}{2x.2x}}-4=-1\)
\(y_{min}=-1\) khi \(x=\dfrac{1}{2}\)
\(y=x+\dfrac{4}{x}\Rightarrow y'=1-\dfrac{4}{x^2}=0\Rightarrow x=-2\)
\(y\left(-2\right)=-4\Rightarrow\max\limits_{x>0}y=-4\) khi \(x=-2\)
HT
9 tháng 2 2021
a/ \(=\lim\limits_{x\rightarrow-\infty}\dfrac{\dfrac{x}{x}+\dfrac{3}{x}}{\dfrac{3x}{x}-\dfrac{1}{x}}=\dfrac{1}{3}\)
b/ \(=\lim\limits_{x\rightarrow-\infty}\dfrac{-\sqrt{\dfrac{x^2}{x^2}-\dfrac{2x}{x^2}+\dfrac{4}{x^2}}-\dfrac{x}{x}}{\dfrac{3x}{x}-\dfrac{1}{x}}=-\dfrac{2}{3}\)
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