Tìm giới hạn :
B = lim \(x\left(\sqrt{x^2+2x}-2\sqrt{x^2+x}+x\right)\) \(\left(x\rightarrow+\infty\right)\)
A. \(+\infty\)
B. \(-\infty\)
C. \(\frac{-1}{4}\)
D. 0
help me !!!!!!
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NV
Nguyễn Việt Lâm
Giáo viên
15 tháng 3 2020
Bài 1:
\(a=\lim\limits_{x\rightarrow-\infty}\frac{2\left|x\right|+1}{3x-1}=\lim\limits_{x\rightarrow-\infty}\frac{-2x+1}{3x-1}=\lim\limits_{x\rightarrow-\infty}\frac{-2+\frac{1}{x}}{3-\frac{1}{x}}=-\frac{2}{3}\)
\(b=\lim\limits_{x\rightarrow+\infty}\frac{\sqrt{9+\frac{1}{x}+\frac{1}{x^2}}-\sqrt{4+\frac{2}{x}+\frac{1}{x^2}}}{1+\frac{1}{x}}=\frac{\sqrt{9}-\sqrt{4}}{1}=1\)
\(c=\lim\limits_{x\rightarrow+\infty}\frac{\sqrt{1+\frac{2}{x}+\frac{3}{x^2}}+4+\frac{1}{x}}{\sqrt{4+\frac{1}{x^2}}+\frac{2}{x}-1}=\frac{1+4}{\sqrt{4}-1}=5\)
\(d=\lim\limits_{x\rightarrow+\infty}\frac{\frac{3}{x}-\frac{2}{x\sqrt{x}}+\sqrt{1-\frac{5}{x^3}}}{2+\frac{4}{x}-\frac{5}{x^2}}=\frac{1}{2}\)
NV
Nguyễn Việt Lâm
Giáo viên
15 tháng 3 2020
Bài 2:
\(a=\lim\limits_{x\rightarrow-\infty}\frac{2+\frac{1}{x}}{1-\frac{1}{x}}=2\)
\(b=\lim\limits_{x\rightarrow-\infty}\frac{2+\frac{3}{x^3}}{1-\frac{2}{x}+\frac{1}{x^3}}=2\)
\(c=\lim\limits_{x\rightarrow+\infty}\frac{x^2\left(3+\frac{1}{x^2}\right)x\left(5+\frac{3}{x}\right)}{x^3\left(2-\frac{1}{x^3}\right)x\left(1+\frac{4}{x}\right)}=\frac{15}{+\infty}=0\)
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 = \(\dfrac{a_1+a_2+...+a_n}{n}\)
19 tháng 2 2023
a: \(\lim\limits_{x->0^-^-}\dfrac{-2x+x}{x\left(x-1\right)}=lim_{x->0^-}\left(\dfrac{-x}{x\left(x-1\right)}\right)\)
\(=lim_{x->0^-}\left(\dfrac{-1}{x-1}\right)=\dfrac{-1}{0-1}=\dfrac{-1}{-1}=1\)
b: \(=lim_{x->-\infty}\left(\dfrac{x^2-x-x^2+1}{\sqrt{x^2-x}+\sqrt{x^2-1}}\right)\)
\(=lim_{x->-\infty}\left(\dfrac{-x+1}{\sqrt{x^2-x}+\sqrt{x^2-1}}\right)\)
\(=lim_{x->-\infty}\left(\dfrac{-1+\dfrac{1}{x}}{-\sqrt{1-\dfrac{1}{x^2}}-\sqrt{1-\dfrac{1}{x^2}}}\right)=\dfrac{-1}{-2}=\dfrac{1}{2}\)
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
4 tháng 12 2023
a: \(\lim\limits_{x\rightarrow+\infty}\dfrac{\sqrt{x^2+x+2}}{x-1}\)
\(=\lim\limits_{x\rightarrow+\infty}\dfrac{\sqrt{1+\dfrac{1}{x}+\dfrac{2}{x^2}}}{1-\dfrac{1}{x}}=\dfrac{\sqrt{1+0+0}}{1-0}\)
\(=\dfrac{1}{1}\)
=1
b: \(\lim\limits_{x\rightarrow-\infty}\left(\sqrt{4x^2-x}+2x\right)\)
\(=\lim\limits_{x\rightarrow-\infty}\dfrac{4x^2-x-4x^2}{\sqrt{4x^2-x}-2x}\)
\(=\lim\limits_{x\rightarrow-\infty}\dfrac{-x}{\sqrt{x^2\left(4-\dfrac{1}{x}\right)}-2x}\)
\(=\lim\limits_{x\rightarrow-\infty}\dfrac{-x}{-x\sqrt{4-\dfrac{1}{x}}-2x}\)
\(=\lim\limits_{x\rightarrow-\infty}\dfrac{1}{\sqrt{4-\dfrac{1}{x}}+2}=\dfrac{1}{\sqrt{4}+2}=\dfrac{1}{2+2}=\dfrac{1}{4}\)
2
18 tháng 11 2023
`a)lim_{x->+oo} (2x-\sqrt{x^2+4x-3})` `ĐK: x < -2-\sqrt{7};x > -2+\sqrt{7}`
`=lim_{x->+oo} [x(2-\sqrt{1+4/x -3/[x^2]}]`
`=+oo`
`b)lim_{x->+oo} (\sqrt{4x^2-3x+1}-2x)`
`=lim_{x->+oo} [4x^2-3x+1-4x^2]/[\sqrt{4x^2-3x+1}+2x]`
`=lim_{x->+oo} [-3x+1]/[\sqrt{4x^2-3x+1}+2x]`
`=lim_{x->+oo} [-3+1/x]/[\sqrt{4-3/x+1/[x^2]}+2]`
`=-3/4`
\(B=\lim\limits_{x\rightarrow+\infty}x\left(\frac{\left(\sqrt{x^2+2x}+x\right)^2-4\left(x^2+x\right)}{\sqrt{x^2+2x}+x+2\sqrt{x^2+x}}\right)\)
\(=\lim\limits_{x\rightarrow+\infty}2x^2\left(\frac{\sqrt{x^2+2x}-x-1}{\sqrt{x^2+2x}+x+2\sqrt{x^2+x}}\right)\)
\(=\lim\limits_{x\rightarrow+\infty}\frac{2x^2\left(x^2+2x-\left(x+1\right)^2\right)}{\left(\sqrt{x^2+2x}+x+2\sqrt{x^2+x}\right)\left(\sqrt{x^2+2x}+x+1\right)}\)
\(=\lim\limits_{x\rightarrow+\infty}\frac{-2x^2}{\left(\sqrt{x^2+2x}+x+2\sqrt{x^2+x}\right)\left(\sqrt{x^2+2x}+x+1\right)}\)
\(=\lim\limits_{x\rightarrow+\infty}\frac{-2x^2}{x^2\left(\sqrt{1+\frac{2}{x}}+1+2\sqrt{1+\frac{1}{x}}\right)\left(\sqrt{1+\frac{2}{x}}+1+\frac{1}{x}\right)}=\frac{-2}{\left(1+1+2\right)\left(1+1+0\right)}=-\frac{1}{4}\)