Question

[Trần Phương Thảo]

bài 1

a. \(\lim\limits_{n\rightarrow+\infty}\left(\sqrt[3]{n^3+3n+1}-n\right)\)

b. \(\lim\limits_{n\rightarrow+\infty}\left(\sqrt[3]{n^3+2n}-\sqrt{n^2+1}\right)\)

c.\(\lim\limits_{n\rightarrow+\infty}\left(\sqrt[3]{n^3+n}-\sqrt[3]{n^3-2n^2}\right)\)

bài 2

a. \(\lim\limits_{n\rightarrow+\infty}\left(2n-\sqrt{n^2+3n}\right)\)

b. \(\lim\limits_{n\rightarrow+\infty}\left(\sqrt{n^2+2}-\sqrt{3n+1}\right)\)

c. \(\lim\limits_{n\rightarrow+\infty}\left(n\sin2n-3n^3\right)\)

Answer 1

[Nguyễn Việt Lâm]

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\(=lim\left[n\left(2-\sqrt{1+\frac{3}{n}}\right)\right]=+\infty\left(2-1\right)=+\infty\)

\(b=lim\left[n\left(\sqrt{1+\frac{2}{n^2}}-\sqrt{\frac{3}{n}+\frac{1}{n^2}}\right)\right]=+\infty\left(1-0\right)=+\infty\)

\(c=lim\left[n^3\left(\frac{sin2n}{n^2}-3\right)\right]=+\infty\left(0-3\right)=-\infty\)

Answer 2

[Ngô Tiến Đạt]

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