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Da nan roi mang meo lam mat het bai -.-
1/ \(=\lim\limits_{x\rightarrow-\infty}\dfrac{\sqrt[3]{\dfrac{3x^3}{x^3}+\dfrac{1}{x^3}}+\sqrt{\dfrac{2x^2}{x^2}+\dfrac{x}{x^2}+\dfrac{1}{x^2}}}{-\sqrt[4]{\dfrac{4x^4}{x^4}+\dfrac{2}{x^4}}}=\dfrac{-\sqrt[3]{3}-\sqrt{2}}{\sqrt[4]{4}}\)
2/ \(=\lim\limits_{x\rightarrow+\infty}\dfrac{8x^7}{\left(-2x^7\right)}=-\dfrac{8}{2^7}\)
3/ \(=\lim\limits_{x\rightarrow+\infty}\dfrac{\left(4x^2-3x+4-4x^2\right)\left(\sqrt{x^2+x+1}+x\right)}{\left(x^2+x+1-x^2\right)\left(\sqrt{4x^2-3x+4}+2x\right)}=\dfrac{-3.2}{2}=-3\)
Xet \(m\ne-3\)
\(=\lim\limits_{x\rightarrow-\infty}x\left(\sqrt[3]{1}+\sqrt{4}+m\right)=x\left(3+m\right)\)
\(=\left[{}\begin{matrix}-\infty\left(m>-3\right)\\+\infty\left(m< -3\right)\end{matrix}\right.\)
Xet \(m=-3\)
\(=\lim\limits_{x\rightarrow-\infty}\left(\sqrt[3]{x^3+2x^2+1}-x-2x-\sqrt{4x^2+2x+3}\right)\)
\(=\lim\limits_{x\rightarrow-\infty}\dfrac{x^3+2x^2+1-x^3}{\sqrt[3]{\left(x^3+2x^2+1\right)^2}+x\sqrt[3]{x^3+2x^2+1}+x^2}-\lim\limits_{x\rightarrow-\infty}\dfrac{4x^2-4x^2-2x-3}{2x-\sqrt{4x^2+2x+3}}\)
\(=\dfrac{2}{3}+\dfrac{1}{2}=\dfrac{7}{6}\)
Bạn bị nhầm số rồi. Xét $m>1; m< 1; m=1$ mới đúng chứ
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
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/ \(=\lim\limits_{x\rightarrow+\infty}x.\left(\dfrac{x^2+2x-x^2}{\sqrt{x^2+2x}+x}+2.\dfrac{x^2-x^2+x}{\sqrt{x^2-x}+x}\right)=+\infty\)
\(\lim\limits_{x\rightarrow-\infty}x^3\left(m+\dfrac{2}{x}\right)\left(\dfrac{m}{x^2}-3\right)=+\infty.\left(3m\right)=-\infty\)
\(\Rightarrow m< 0\Rightarrow\) có 20 giá trị nguyên của m
\(=\lim\limits_{x\rightarrow-\infty}\dfrac{1-\sqrt{4x^2-x+5}}{-ax+2}=\lim\limits_{x\rightarrow-\infty}\dfrac{\dfrac{1}{x}+\sqrt{4-\dfrac{1}{x}+\dfrac{5}{x^2}}}{-a+\dfrac{2}{x}}=\dfrac{2}{-a}=\dfrac{2}{3}\)
\(\Rightarrow a=-3\)
\(\lim\limits_{x\rightarrow-2}\dfrac{x^3+2x^2}{\sqrt{x^2+4x+4}}=\lim\limits_{x\rightarrow-2}\dfrac{x^2\left(x+2\right)}{\sqrt{\left(x+2\right)^2}}\)
\(=\lim\limits_{x\rightarrow-2}x^2=\left(-2\right)^2=4\)
p/s: bài này mình chưa học trên lớp nên ko chắc 100% đúng
\(\lim\limits_{x\rightarrow+\infty}\dfrac{\sqrt{x+1}}{\sqrt{x+\sqrt{x+1}}+\sqrt{x}}=\lim\limits_{x\rightarrow+\infty}\dfrac{\sqrt{1+\dfrac{1}{x}}}{\sqrt{1+\sqrt{\dfrac{1}{x}+\dfrac{1}{x^2}}}+1}=\dfrac{1}{1+1}=\dfrac{1}{2}\)
Câu c số 1 trong hay ngoài căn nhỉ?
`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`
\(\lim\limits_{x\rightarrow-\infty}\dfrac{\left(2x^2-1\right)\left(mx+3\right)}{x^3+4x+7}=\lim\limits_{x\rightarrow-\infty}\dfrac{\left(2-\dfrac{1}{x^2}\right)\left(m+\dfrac{3}{x}\right)}{1+\dfrac{4}{x^2}+\dfrac{7}{x^3}}=2m\)
\(\Rightarrow2m=6\Rightarrow m=3\)