tìm a để giới hạn : \(lim_{x->+\infty}\left(\sqrt{x^2-3x+5}+ax\right)=+\infty\)
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12 giờ trước (21:50)
\(\lim_{x\to\infty}\left(\frac{1}{\sqrt[3]{n^3+1}-n}\right)=\lim_{x\to\infty}\left(\frac{1}{\frac{n^3+1-n^3}{\sqrt[3]{\left(n^3+1\right)^2}+n\cdot\sqrt[3]{n^3+1}+n^2}}\right)\)
\(=\lim_{x\to\infty}\left(\sqrt[3]{\left(n^3+1\right)^2}+n\cdot\sqrt[3]{n^3+1}+n^2\right)=\lim_{x\to\infty}\left\lbrack n^2\left(\sqrt[3]{\left(1+\frac{1}{n^3}\right)^2}+\sqrt[3]{1+\frac{1}{n^3}}+1\right)\right\rbrack=+\infty\)
\(\lim_{x\to\infty}\left(\sqrt[3]{n^3-2n^2}-n\right)\)
\(=\lim_{x\to\infty}\frac{n^3-2n^2-n^3}{\sqrt[3]{\left(n^3-2n^2\right)^2}+n\cdot\sqrt[3]{n^3-2n^2}+n^2}\)
\(=\lim_{x\to\infty}\frac{-2n^2}{n^2\cdot\left\lbrack\sqrt[3]{\left(1-\frac{2}{n}\right)^2}+\sqrt[3]{1-\frac{2}{n}}+1\right\rbrack}=\lim_{x\to\infty}\frac{-2}{\left\lbrack\sqrt[3]{\left(1-\frac{2}{n}\right)^2}+\sqrt[3]{1-\frac{2}{n}}+1\right\rbrack}\)
=-∞
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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\(\lim\limits_{x\to 2019}\frac{\sqrt{x+285}-48}{\sqrt{x-2018}-\sqrt{2020-x}}=\lim\limits_{x\to -\infty}(\sqrt{x+285}-48).\lim\limits_{x\to -\infty}\frac{1}{\sqrt{x-2018}-\sqrt{2020-x}}\)
\(=\lim\limits_{x\to 2019}\frac{x-2019}{\sqrt{x+285}+48}.\lim\limits_{x\to 2019}\frac{\sqrt{x-2018}+\sqrt{2020-x}}{2(x-2019)}=\lim\limits_{x\to 2019}\frac{\sqrt{x-2018}+\sqrt{2020-x}}{2(\sqrt{x+285}+48)}=\frac{1}{96}\)
HT
18 tháng 2 2021
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
9 tháng 2 2021
\(\lim\limits_{x\rightarrow+\infty}\left(\sqrt[n]{\left(x+a_1\right)\left(x+a_2\right)...\left(x+a_n\right)}-x\right)\\ =\lim\limits_{x\rightarrow+\infty}\left(\dfrac{\left(x+a_1\right)\left(x+a_2\right)...\left(x+a_n\right)-x^n}{\sqrt[n]{\left(\left(x+a_1\right)\left(x+a_2\right)...\left(x+a_n\right)\right)^{n-1}}+...+x^{n-1}}\right)\)
= hệ số xn-1 trên tử/hệ số xn-1 dưới mẫu = \(\dfrac{a_1+a_2+...+a_n}{n}\)
TH
1
9 tháng 2 2021
\(\lim\limits_{x\rightarrow+\infty}=\left(\sqrt[3]{x^3+3\text{x}^2}-\sqrt{x^2-2\text{x}}\right)\\ =\lim\limits_{x\rightarrow+\infty}\left(\sqrt[3]{x^3+3\text{x}^2}-x+x-\sqrt{x^2-2x}\right)\\ =\lim\limits_{x\rightarrow+\infty}\left(\dfrac{3\text{x}^2}{\sqrt[3]{\left(x^3+3\text{x}^2\right)^2}+x\sqrt[3]{x^3+3\text{x}^2}+x^2}+\dfrac{2\text{x}}{x+\sqrt{x^2-2x}}\right)\\ =\dfrac{3}{1+1+1}+\dfrac{2}{1+1}=2\)
Bạn tự hiểu là giới hạn khi x tiến tới dương vô cực
\(=lim\left[x\left(\sqrt{1-\frac{3}{x}+\frac{5}{x^2}}+a\right)\right]=lim\left[x\left(1-a\right)\right]\)
Do \(x\rightarrow+\infty\) nên để giới hạn đã cho bằng \(+\infty\Leftrightarrow1-a>0\Rightarrow a< 1\)
thanks