\(lim_{x->+\infty}\left(\sqrt{x^2+2x}-2\sqrt{x^2+x}+x\right)\)
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AH
Akai Haruma
Giáo viên
20 tháng 3 2020
Lời giải:
\(\lim\limits_{x\to \pm\infty}\sqrt{x^2-3x+4}=\lim\limits_{x\to \pm\infty}\sqrt{x^2}.\lim\limits_{x\to \pm \infty}\sqrt{1-\frac{3}{x}+\frac{4}{x^2}}=\lim\limits_{x\to \pm\infty}|x|.1=+\infty \)
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\(\lim\limits_{x\to +\infty}x(\sqrt{x^2+5}+x)=\lim\limits_{x\to +\infty}x^2.\lim\limits_{x\to +\infty}(\sqrt{1+\frac{5}{x^2}}+1)=2(+\infty )=+\infty \)
\(\lim\limits_{x\to -\infty}x(\sqrt{x^2+5}+x)=\lim\limits_{x\to -\infty}\frac{5x}{\sqrt{x^2+5}-x}=\lim\limits_{x\to -\infty}\frac{-5}{\sqrt{1+\frac{5}{x^2}}+1}=\frac{-5}{2}\)
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\(\lim\limits_{x\to 2019}\frac{\sqrt{x+285}-48}{\sqrt{x-2018}-\sqrt{2020-x}}=\lim\limits_{x\to -\infty}(\sqrt{x+285}-48).\lim\limits_{x\to -\infty}\frac{1}{\sqrt{x-2018}-\sqrt{2020-x}}\)
\(=\lim\limits_{x\to 2019}\frac{x-2019}{\sqrt{x+285}+48}.\lim\limits_{x\to 2019}\frac{\sqrt{x-2018}+\sqrt{2020-x}}{2(x-2019)}=\lim\limits_{x\to 2019}\frac{\sqrt{x-2018}+\sqrt{2020-x}}{2(\sqrt{x+285}+48)}=\frac{1}{96}\)
AH
Akai Haruma
Giáo viên
16 tháng 3 2020
Lời giải:
\(\lim\limits_{x\to \pm\infty}\sqrt{x^2-3x+4}=\lim\limits_{x\to \pm\infty}\sqrt{x^2}.\lim\limits_{x\to \pm \infty}\sqrt{1-\frac{3}{x}+\frac{4}{x^2}}=\lim\limits_{x\to \pm\infty}|x|.1=+\infty \)
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\(\lim\limits_{x\to +\infty}x(\sqrt{x^2+5}+x)=\lim\limits_{x\to +\infty}x^2.\lim\limits_{x\to +\infty}(\sqrt{1+\frac{5}{x^2}}+1)=2(+\infty )=+\infty \)
\(\lim\limits_{x\to -\infty}x(\sqrt{x^2+5}+x)=\lim\limits_{x\to -\infty}\frac{5x}{\sqrt{x^2+5}-x}=\lim\limits_{x\to -\infty}\frac{-5}{\sqrt{1+\frac{5}{x^2}}+1}=\frac{-5}{2}\)
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\(\lim\limits_{x\to 2019}\frac{\sqrt{x+285}-48}{\sqrt{x-2018}-\sqrt{2020-x}}=\lim\limits_{x\to -\infty}(\sqrt{x+285}-48).\lim\limits_{x\to -\infty}\frac{1}{\sqrt{x-2018}-\sqrt{2020-x}}\)
\(=\lim\limits_{x\to 2019}\frac{x-2019}{\sqrt{x+285}+48}.\lim\limits_{x\to 2019}\frac{\sqrt{x-2018}+\sqrt{2020-x}}{2(x-2019)}=\lim\limits_{x\to 2019}\frac{\sqrt{x-2018}+\sqrt{2020-x}}{2(\sqrt{x+285}+48)}=\frac{1}{96}\)
24 tháng 11 2023
\(\lim\limits_{x\rightarrow-\infty}\left(\sqrt{x^2+1}+x-1\right)\)
\(=\lim\limits_{x\rightarrow-\infty}\dfrac{x^2+1-\left(x-1\right)^2}{\sqrt{x^2+1}-x+1}\)
\(=\lim\limits_{x\rightarrow-\infty}\dfrac{x^2+1-x^2+2x-1}{-x\sqrt{1+\dfrac{1}{x^2}}-x+1}\)
\(=\lim\limits_{x\rightarrow-\infty}\dfrac{-2x}{x\left(-\sqrt{1+\dfrac{1}{x^2}}-1+\dfrac{1}{x}\right)}\)
\(=\lim\limits_{x\rightarrow-\infty}\dfrac{-2}{-\sqrt{1+\dfrac{1}{x^2}}-1+\dfrac{1}{x}}\)
\(=\dfrac{-2}{-\sqrt{1+0}-1+0}=\dfrac{-2}{-1-1}=1\)
b: \(\lim\limits\dfrac{\sqrt{4n^2+n-1}+n}{\sqrt{n^4+2n^3-1}-n}\)
\(=\lim\limits\dfrac{n\left(\sqrt{4+\dfrac{1}{n}-\dfrac{1}{n^2}}+1\right)}{n^2\cdot\sqrt{1+\dfrac{2}{n}-\dfrac{1}{n^4}}-n^2\cdot\dfrac{1}{n}}\)
\(=\lim\limits\dfrac{n\left(\sqrt{4+\dfrac{1}{n}-\dfrac{1}{n^2}}+1\right)}{n^2\left(\sqrt{1+\dfrac{2}{n}-\dfrac{1}{n^4}}-\dfrac{1}{n}\right)}\)
\(=\lim\limits\dfrac{\sqrt{4+\dfrac{1}{n}-\dfrac{1}{n^2}}+1}{n\left(\sqrt{1+\dfrac{2}{n}-\dfrac{1}{n^4}}-\dfrac{1}{n}\right)}\)
\(=\lim\limits\dfrac{\sqrt{\dfrac{4}{n^2}+\dfrac{1}{n^3}-\dfrac{1}{n^4}}+\dfrac{1}{n}}{\sqrt{1+\dfrac{2}{n}-\dfrac{1}{n^4}}-\dfrac{1}{n}}\)
\(=\dfrac{0}{\sqrt{1+0-0}-0}=\dfrac{0}{1}=0\)
NM
1
12 tháng 2 2020
\(=\lim\limits_{x\rightarrow+\infty}\frac{x+\sqrt{x+\sqrt{x}}-x}{\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}}\)
\(=\lim\limits_{x\rightarrow+\infty}\frac{\sqrt{x+\sqrt{x}}}{\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}}\)
\(=\lim\limits_{x\rightarrow+\infty}\frac{\sqrt{1+\sqrt{\frac{1}{x}}}}{\sqrt{1+\sqrt{\frac{1}{x}+\sqrt{\frac{1}{x^3}}}}+1}=\frac{1}{1+1}=\frac{1}{2}\)
HT
9 tháng 2 2021
1/ \(=\lim\limits_{x\rightarrow+\infty}x\left(\sqrt{\dfrac{x^2}{x^2}+\dfrac{x}{x^2}+\dfrac{1}{x^2}}-\sqrt[3]{\dfrac{2x^3}{x^3}+\dfrac{x}{x^3}-\dfrac{1}{x^3}}\right)=x\left(1-\sqrt[3]{2}\right)=-\infty\)
2/ \(=\lim\limits_{x\rightarrow+\infty}\dfrac{4x^2+x+1-4x^2}{\sqrt{4x^2+x+1}+2x}=\lim\limits_{x\rightarrow+\infty}\dfrac{\dfrac{x}{x}+\dfrac{1}{x}}{\sqrt{\dfrac{4x^2}{x^2}+\dfrac{x}{x^2}+\dfrac{1}{x^2}}+\dfrac{2x}{x}}=\dfrac{1}{2+2}=\dfrac{1}{4}\)
3/ \(=\lim\limits_{x\rightarrow-\infty}\dfrac{x^3+x^2+1-x^3}{\left(\sqrt[3]{x^3+x^2+1}\right)^2+x.\sqrt[3]{x^3+x^2+1}+x^2}+\dfrac{x^2+x+1-x^2}{\sqrt{x^2+x+1}-x}\)
\(=\lim\limits_{x\rightarrow-\infty}\dfrac{\dfrac{x^2}{x^2}+\dfrac{1}{x^2}}{\dfrac{\left(\sqrt[3]{x^3+x^2+1}\right)^2}{x^2}+\dfrac{x}{x^2}\sqrt[3]{x^3+x^2+1}+\dfrac{x^2}{x^2}}+\dfrac{\dfrac{x}{x}+\dfrac{1}{x}}{-\sqrt{\dfrac{x^2}{x^2}+\dfrac{x}{x^2}+\dfrac{1}{x^2}}-\dfrac{x}{x}}=\dfrac{1}{3}-\dfrac{1}{2}=-\dfrac{1}{6}\)
4/ \(=\lim\limits_{x\rightarrow+\infty}\left(\sqrt{x^2+x+1}-x\right)+\lim\limits_{x\rightarrow+\infty}2\left(x-\sqrt{x^2-x}\right)\)
\(=\lim\limits_{x\rightarrow+\infty}\dfrac{x^2+x+1-x^2}{\sqrt{x^2+x+1}+x}+\lim\limits_{x\rightarrow+\infty}2.\dfrac{x^2-x^2+x}{x+\sqrt{x^2-x}}\)
\(=\lim\limits_{x\rightarrow+\infty}\dfrac{\dfrac{x}{x}+\dfrac{1}{x}}{\sqrt{\dfrac{x^2}{x^2}+\dfrac{x}{x^2}+\dfrac{1}{x^2}}+\dfrac{x}{x}}+\lim\limits_{x\rightarrow+\infty}2.\dfrac{\dfrac{x}{x}}{\dfrac{x}{x}+\sqrt{\dfrac{x^2}{x^2}-\dfrac{x}{x^2}}}=\dfrac{1}{2}+\dfrac{2}{2}=\dfrac{3}{2}\)
5/ \(=\lim\limits_{x\rightarrow+\infty}x.\left(\dfrac{x^2+2x-x^2}{\sqrt{x^2+2x}+x}+2.\dfrac{x^2-x^2+x}{\sqrt{x^2-x}+x}\right)=+\infty\)
NN
0
HT
9 tháng 2 2021
1/ \(=\lim\limits_{x\rightarrow-\infty}\dfrac{x^2-x^2-x-x}{x+\sqrt{x^2+x+1}}=\dfrac{-2}{1-1}=-\infty\)
2/ tien toi +- vo cung?
3/ \(=\lim\limits_{x\rightarrow+\infty}\dfrac{8x^3+2x-8x^3}{\sqrt[3]{\left(8x^3+2x\right)^2}+2x.\sqrt[3]{8x^3+2x}+4x^2}=\dfrac{\dfrac{2x}{x^2}}{\dfrac{4x^2}{x^2}+\dfrac{4x^2}{x^2}+\dfrac{4x^2}{x^2}}=0\)
4/ \(\lim\limits_{x\rightarrow+\infty}\dfrac{16x^4+3x+1-16x^4}{\sqrt[4]{\left(16x^4+3x+1\right)^3}+2x.\sqrt[4]{\left(16x^4+3x+1\right)^2}+4x^2.\sqrt[4]{16x^4+3x+1}+8x^3}+\lim\limits_{x\rightarrow+\infty}\dfrac{4x^2-4x^2-2}{2x+\sqrt{4x^2+2}}=\dfrac{\dfrac{3x}{x^3}}{8+8+8+8}-\dfrac{\dfrac{2}{x}}{2+2}=0\)
5/ \(=\lim\limits_{x\rightarrow+\infty}\dfrac{x^2+1-x^2}{\sqrt{x^2+1}+x}+\lim\limits_{x\rightarrow+\infty}\dfrac{x^2-x-x^2}{\sqrt{x^2-x}+x}=\dfrac{\dfrac{1}{x}}{1+1}-\dfrac{\dfrac{x}{x}}{1+1}=-\dfrac{1}{2}\)
\(=\lim\limits_{x\rightarrow+\infty}\left(\sqrt{x^2+2x}-\sqrt{x^2+x}+x-\sqrt{x^2+x}\right)\)
\(=\lim\limits_{x\rightarrow+\infty}\left(\dfrac{x}{\sqrt{x^2+2x}+\sqrt{x^2+x}}-\dfrac{x}{x+\sqrt{x^2+x}}\right)\)
\(=\lim\limits_{x\rightarrow+\infty}\left(\dfrac{1}{\sqrt{1+\dfrac{2}{x}}+\sqrt{1+\dfrac{1}{x}}}-\dfrac{1}{1+\sqrt{1+\dfrac{1}{x}}}\right)=\dfrac{1}{2}-\dfrac{1}{2}=0\)
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