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Vì cả 3 giới hạn kia đều ko tồn tại, chỉ có giới hạn cuối là tồn tại (do hàm sin, cos là hàm tuần hoàn có chu kì, do đó giới hạn vô cực ko tồn tại)
Đặt \(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+....+\frac{1}{n\left(n+1\right)}=A\)
\(\Leftrightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{n}-\frac{1}{n+1}\)
\(\Leftrightarrow A=\frac{n+1}{n+1}-\frac{1}{n+1}=\frac{n}{n+1}\)
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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\(\lim\limits_{x\to 2019}\frac{\sqrt{x+285}-48}{\sqrt{x-2018}-\sqrt{2020-x}}=\lim\limits_{x\to -\infty}(\sqrt{x+285}-48).\lim\limits_{x\to -\infty}\frac{1}{\sqrt{x-2018}-\sqrt{2020-x}}\)
\(=\lim\limits_{x\to 2019}\frac{x-2019}{\sqrt{x+285}+48}.\lim\limits_{x\to 2019}\frac{\sqrt{x-2018}+\sqrt{2020-x}}{2(x-2019)}=\lim\limits_{x\to 2019}\frac{\sqrt{x-2018}+\sqrt{2020-x}}{2(\sqrt{x+285}+48)}=\frac{1}{96}\)
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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\(\lim\limits_{x\to 2019}\frac{\sqrt{x+285}-48}{\sqrt{x-2018}-\sqrt{2020-x}}=\lim\limits_{x\to -\infty}(\sqrt{x+285}-48).\lim\limits_{x\to -\infty}\frac{1}{\sqrt{x-2018}-\sqrt{2020-x}}\)
\(=\lim\limits_{x\to 2019}\frac{x-2019}{\sqrt{x+285}+48}.\lim\limits_{x\to 2019}\frac{\sqrt{x-2018}+\sqrt{2020-x}}{2(x-2019)}=\lim\limits_{x\to 2019}\frac{\sqrt{x-2018}+\sqrt{2020-x}}{2(\sqrt{x+285}+48)}=\frac{1}{96}\)
Giới hạn của dãy nên bạn tự hiểu n tiến tới dương vô cực
1.
\(lim\frac{3n+1}{\sqrt[3]{\left(n^3+3n+1\right)^2}+n\sqrt{n^3+3n+1}+n^2}=lim\frac{3+\frac{1}{n}}{\sqrt[3]{\frac{\left(n^3+3n+1\right)^2}{n^3}}+\sqrt{n^3+3n+1}+n}=\frac{3}{\infty}=0\)
b=\(lim\left(\sqrt[3]{n^3+2n}-n+n-\sqrt{n^2+1}\right)=lim\left(\frac{2n}{\sqrt[3]{\left(n^3+2n\right)^2}+n\sqrt[3]{n^3+2n}+n^2}-\frac{1}{n+\sqrt{n^2+1}}\right)\)
\(=lim\left(\frac{2}{\sqrt[3]{\frac{\left(n^3+2n\right)^2}{n^3}}+\sqrt[3]{n^3+2n}+n}-\frac{1}{n+\sqrt{n^2+1}}\right)=0-0=0\)
c\(=lim\left(\frac{2n^2+n}{\sqrt[3]{\left(n^3+n\right)^2}+\sqrt[3]{\left(n^3+n\right)\left(n^3-2n^2\right)}+\sqrt[3]{\left(n^3-2n^2\right)^2}}\right)\)
\(=lim\left(\frac{2+\frac{1}{n}}{\sqrt[3]{\left(1+\frac{1}{n^2}\right)^2}+\sqrt[3]{\left(1+\frac{1}{n^2}\right)\left(1-\frac{2}{n}\right)}+\sqrt[3]{\left(1-\frac{2}{n}\right)^2}}\right)=\frac{2}{1+1.1+1}=\frac{2}{3}\)
2.
a\(=lim\left[n\left(2-\sqrt{1+\frac{3}{n}}\right)\right]=+\infty\left(2-1\right)=+\infty\)
\(b=lim\left[n\left(\sqrt{1+\frac{2}{n^2}}-\sqrt{\frac{3}{n}+\frac{1}{n^2}}\right)\right]=+\infty\left(1-0\right)=+\infty\)
\(c=lim\left[n^3\left(\frac{sin2n}{n^2}-3\right)\right]=+\infty\left(0-3\right)=-\infty\)
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 \(+\infty\Leftrightarrow1-a>0\Rightarrow a< 1\)
\(\lim_{x\to\infty}\left(\frac{1}{\sqrt[3]{n^3+1}-n}\right)=\lim_{x\to\infty}\left(\frac{1}{\frac{n^3+1-n^3}{\sqrt[3]{\left(n^3+1\right)^2}+n\cdot\sqrt[3]{n^3+1}+n^2}}\right)\)
\(=\lim_{x\to\infty}\left(\sqrt[3]{\left(n^3+1\right)^2}+n\cdot\sqrt[3]{n^3+1}+n^2\right)=\lim_{x\to\infty}\left\lbrack n^2\left(\sqrt[3]{\left(1+\frac{1}{n^3}\right)^2}+\sqrt[3]{1+\frac{1}{n^3}}+1\right)\right\rbrack=+\infty\)
\(\lim_{x\to\infty}\left(\sqrt[3]{n^3-2n^2}-n\right)\)
\(=\lim_{x\to\infty}\frac{n^3-2n^2-n^3}{\sqrt[3]{\left(n^3-2n^2\right)^2}+n\cdot\sqrt[3]{n^3-2n^2}+n^2}\)
\(=\lim_{x\to\infty}\frac{-2n^2}{n^2\cdot\left\lbrack\sqrt[3]{\left(1-\frac{2}{n}\right)^2}+\sqrt[3]{1-\frac{2}{n}}+1\right\rbrack}=\lim_{x\to\infty}\frac{-2}{\left\lbrack\sqrt[3]{\left(1-\frac{2}{n}\right)^2}+\sqrt[3]{1-\frac{2}{n}}+1\right\rbrack}\)
=-∞