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l i m v n = 0 ⇒ | v n | có thể nhỏ hơn một số dương bé tuỳ ý, kể từ một số hạng nào đó trở đi (1)
Vì | u n | ≤ v n v à v n ≤ | v n | với mọi n, nên | u n | ≤ | v n | với mọi n. (2)
Từ (1) và (2) suy ra | u n | cũng có thể nhỏ hơn một số dương bé tuỳ ý, kể từ một số hạng nào đó trở đi, nghĩa là l i m u n = 0
`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`
Hic nan qua :( Lam vay
P/s: Anh Lam check all ho em nhung bai em lam nhe :( Em cam on
1/ \(=\lim\limits_{x\rightarrow+\infty}\dfrac{x^2-x+1-x^2}{\sqrt{x^2-x+1}+x}=\dfrac{-1}{1+1}=-\dfrac{1}{2}\)
2/ \(=\lim\limits_{x\rightarrow-\infty}x\left(\dfrac{4x^2+1-x^2}{\sqrt{4x^2+1}+x}\right)=\lim\limits_{x\rightarrow-\infty}\dfrac{\dfrac{x}{x}}{-\sqrt{\dfrac{4x^2}{x^2}+\dfrac{1}{x^2}}+\dfrac{x}{x}}=\dfrac{1}{-2+1}=-1\)
3/ \(=\lim\limits_{x\rightarrow-\infty}x^5\left(4-\dfrac{3}{x^2}+\dfrac{1}{x^4}+\dfrac{1}{x^5}\right)=-\infty\)
4/ \(=\lim\limits_{x\rightarrow+\infty}\sqrt{x^4}\left(\sqrt{1-\dfrac{x^3}{x^4}+\dfrac{x^2}{x^4}-\dfrac{x}{x^4}}\right)=+\infty\)
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/ \(=\lim\limits_{x\rightarrow+\infty}\dfrac{x^2+1-x^2}{\sqrt{x^2+1}+x}+\lim\limits_{x\rightarrow+\infty}\dfrac{x^2-x-x^2}{\sqrt{x^2-x}+x}=\dfrac{\dfrac{1}{x}}{1+1}-\dfrac{\dfrac{x}{x}}{1+1}=-\dfrac{1}{2}\)
1) \(\lim\limits_{n\rightarrow\infty}\dfrac{3n^2+5n-3}{-n+5}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{n\left(3n+5-\dfrac{3}{n}\right)}{-n\left(1-\dfrac{5}{n}\right)}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{3n+5-\dfrac{3}{n}}{-\left(1-\dfrac{5}{n}\right)}\)
\(=\left[{}\begin{matrix}-\infty\left(n\rightarrow+\infty\right)\\+\infty\left(n\rightarrow-\infty\right)\end{matrix}\right.\)
Bài 2,3 tương tự, bạn tự làm nhé!
a/ \(=\lim\limits_{x\rightarrow-\infty}x^3\left(3+\dfrac{5x^2}{x^3}-\dfrac{9\sqrt{2}x}{x^3}-\dfrac{2017}{x^3}\right)=3.x^3=-\infty\)
b/ \(=\lim\limits_{x\rightarrow+\infty}x\left(\sqrt{1+\dfrac{x}{x^2}+\dfrac{1}{x^2}}-\sqrt[3]{2+\dfrac{x}{x^3}-\dfrac{1}{x^3}}\right)=\left(1-\sqrt[3]{2}\right)x=-\infty\)
c/ \(=\lim\limits_{x\rightarrow-\infty}\dfrac{x^2-x^2-x-1}{x+\sqrt{x^2+x+1}}=\lim\limits_{x\rightarrow-\infty}\dfrac{-\dfrac{x}{x}-\dfrac{1}{x}}{\dfrac{x}{x}-\sqrt{\dfrac{x^2}{x^2}+\dfrac{x}{x^2}+\dfrac{1}{x^2}}}=-\dfrac{1}{1-1}=-\infty\)
d/ \(=\lim\limits_{x\rightarrow-\infty}\left(\sqrt[3]{x^3+x^2+1}-x\right)+\lim\limits_{x\rightarrow-\infty}\left(x+\sqrt{x^2+x+1}\right)\)
\(=\lim\limits_{x\rightarrow-\infty}\dfrac{x^3+x^2+1-x^3}{\left(\sqrt[3]{x^3+x^2+1}\right)^2+x\sqrt[3]{x^3+x^2+1}-x^2}+\lim\limits_{x\rightarrow-\infty}\dfrac{x^2-x^2-x-1}{x-\sqrt{x^2+x+1}}\)
\(=\lim\limits_{x\rightarrow-\infty}\dfrac{x^2+1}{\left(-x\sqrt[3]{\dfrac{x^3}{x^3}+\dfrac{x^2}{x^3}+\dfrac{1}{x^3}}\right)^2-x.x\sqrt[3]{\dfrac{x^3}{x^3}+\dfrac{x^2}{x^3}+\dfrac{1}{x^3}}-x^2}+\lim\limits_{x\rightarrow-\infty}\dfrac{-x-1}{x+x\sqrt{\dfrac{x^2}{x^2}+\dfrac{x}{x^2}+\dfrac{1}{x^2}}}\)
\(=\dfrac{1}{1-1-1}+\dfrac{-1}{1+1}=-1-\dfrac{1}{2}=-\dfrac{3}{2}\)
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/ \(=\lim\limits_{x\rightarrow+\infty}x.\left(\dfrac{x^2+2x-x^2}{\sqrt{x^2+2x}+x}+2.\dfrac{x^2-x^2+x}{\sqrt{x^2-x}+x}\right)=+\infty\)
1) \(\lim\limits_{n\rightarrow\infty}\dfrac{6n-8}{n-1}=\lim\limits_{n\rightarrow\infty}\dfrac{2n\left(1-\dfrac{4}{n}\right)}{n\left(1-\dfrac{1}{n}\right)}=2\)
2) \(\lim\limits_{n\rightarrow\infty}\dfrac{n^2+5n-3}{4n^3-2n+5}=\lim\limits_{n\rightarrow\infty}\dfrac{n^2\left(1+\dfrac{5}{n}-\dfrac{3}{n^2}\right)}{n^3\left(4-\dfrac{2}{n^2}+\dfrac{5}{n^3}\right)}=\dfrac{1}{4n}=\infty\)
3) \(\lim\limits_{n\rightarrow\infty}\left(-2n^5+4n^4-3n^2+4\right)=\lim\limits_{n\rightarrow\infty}n^5\left(-2+\dfrac{4}{n}-\dfrac{3}{n^2}+\dfrac{4}{n^5}\right)=-2n^5=-\infty\)
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/ \(\lim\limits_{x\rightarrow\pm\infty}\dfrac{\dfrac{2x^2}{x^2}-\dfrac{1}{x^2}}{\dfrac{3}{x^2}-\dfrac{x^2}{x^2}}=\dfrac{2}{-1}=-2\)
\(a=\lim\limits_{x\rightarrow+\infty}\frac{x+\frac{8}{x^2}}{1+\frac{2}{x}+\frac{1}{x^2}+\frac{2}{x^3}}=\frac{+\infty}{1}=+\infty\)
\(b=\lim\limits_{x\rightarrow-\infty}\frac{x\left(\frac{1}{x}+3\right)}{\left|x\right|\sqrt{2+\frac{3}{x^2}}}=\lim\limits_{x\rightarrow-\infty}\frac{x\left(\frac{1}{x}+3\right)}{-x\sqrt{2+\frac{3}{x^2}}}=\frac{3}{-\sqrt{2}}=\frac{-3\sqrt{2}}{2}\)
\(c=\lim\limits_{x\rightarrow-\infty}\frac{x^2\sqrt[3]{\frac{1}{x^6}+\frac{1}{x^2}+1}}{x^2\sqrt{\frac{1}{x^2}+\frac{1}{x}+1}}=\frac{1}{1}=1\)