\(\lim\limits_{x\rightarrow-\infty}\dfrac{\sqrt{4x^2-2}+\sqrt[3]{x^3+1}}{\sqrt{x^2+1}-x}\)
\(\lim\limits_{x\rightarrow-\infty}\dfrac{2x+3}{\sqrt{2x^2-3}}\)
\(\lim\limits_{x\rightarrow\pm\infty}\dfrac{2x^2-1}{3-x^2}\)
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Những câu hỏi liên quan
HT
9 tháng 2 2021
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/ \(=\lim\limits_{x\rightarrow+\infty}\dfrac{\left(4x^2-3x+4-4x^2\right)\left(\sqrt{x^2+x+1}+x\right)}{\left(x^2+x+1-x^2\right)\left(\sqrt{4x^2-3x+4}+2x\right)}=\dfrac{-3.2}{2}=-3\)
HT
9 tháng 2 2021
1/ \(=\lim\limits_{x\rightarrow+\infty}\dfrac{\dfrac{2x}{x}-\sqrt{\dfrac{3x^2}{x^2}+\dfrac{2}{x^2}}}{\dfrac{5x}{x}+\sqrt{\dfrac{x^2}{x^2}+\dfrac{1}{x^2}}}=\dfrac{2-\sqrt{3}}{5+1}=\dfrac{2-\sqrt{3}}{6}\)
2/ \(=\lim\limits_{x\rightarrow+\infty}\sqrt{\dfrac{\dfrac{x^2}{x^4}+\dfrac{1}{x^4}}{\dfrac{2x^4}{x^4}+\dfrac{x^2}{x^4}-\dfrac{3}{x^4}}}=0\)
3/ \(=\lim\limits_{x\rightarrow-\infty}\dfrac{-\sqrt[3]{\dfrac{x^6}{x^6}+\dfrac{x^4}{x^6}+\dfrac{1}{x^6}}}{\sqrt{\dfrac{x^4}{x^4}+\dfrac{x^3}{x^4}+\dfrac{1}{x^4}}}=-1\)
HT
27 tháng 1 2021
a/ \(\lim\limits_{x\rightarrow+\infty}\dfrac{\dfrac{x}{x}\sqrt{x^2+1}+\dfrac{2x}{x}+\dfrac{1}{x}}{\dfrac{x}{x}\sqrt[3]{\dfrac{2x^3}{x^3}+\dfrac{x}{x^3}+\dfrac{1}{x^3}}+\dfrac{x}{x}}=\lim\limits_{x\rightarrow+\infty}\dfrac{\sqrt{x^2+1}+2}{\sqrt[3]{2}+1}=+\infty\)
b/ \(=\lim\limits_{x\rightarrow1}\dfrac{\sqrt{2.1^2-1+1}-\sqrt[3]{2.1+3}}{3.1^2-2}=...\)
c/ \(\lim\limits_{x\rightarrow+\infty}\dfrac{x\sqrt{\dfrac{4x^2}{x^2}+\dfrac{x}{x^2}}+x\sqrt[3]{\dfrac{8x^3}{x^3}+\dfrac{x}{x^3}-\dfrac{1}{x^3}}}{x\sqrt[4]{\dfrac{x^4}{x^4}+\dfrac{3}{x^4}}}=\dfrac{2+2}{1}=4\)
HT
9 tháng 2 2021
1/ \(=\lim\limits_{x\rightarrow-\infty}\dfrac{-\sqrt{\dfrac{4x^2}{x^2}-\dfrac{2}{x^2}}+\sqrt[3]{\dfrac{x^3}{x^3}+\dfrac{1}{x^3}}}{-\sqrt{\dfrac{x^2}{x^2}+\dfrac{1}{x^2}}-\dfrac{x}{x}}=\dfrac{-1+1}{-1-1}=0\)
2/ \(=\lim\limits_{x\rightarrow-\infty}\dfrac{\sqrt{x^2+1}+2+\dfrac{1}{x}}{\sqrt[3]{\dfrac{2x^3}{x^3}+\dfrac{x}{x^3}+\dfrac{1}{x^3}}+\dfrac{x}{x}}=\dfrac{+\infty+2}{\sqrt[3]{2}+1}=+\infty\)
AH
Akai Haruma
Giáo viên
14 tháng 5 2021
Lời giải:
a)
\(\lim\limits_{x\to +\infty}\frac{\sqrt[3]{x^3+2x^2-4x+1}}{\sqrt{2x^2+x-8}}=\lim\limits_{x\to +\infty}\frac{\sqrt[3]{1+\frac{2}{x}-\frac{4}{x^2}+\frac{1}{x^3}}}{\sqrt{2+\frac{1}{x}-\frac{8}{x^2}}}\)
\(=\frac{1}{\sqrt{2}}\)
b)
\(\lim\limits_{x\to -\infty}\frac{\sqrt{x^2-2x+4}-x}{3x-1}=\lim\limits_{x\to -\infty}\frac{\sqrt{1-\frac{2}{x}+\frac{4}{x^2}}+1}{-3+\frac{1}{x}}=\frac{-1}{3}\)
HT
9 tháng 2 2021
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/ \(=\lim\limits_{x\rightarrow+\infty}\dfrac{x^2+1-x^2}{\sqrt{x^2+1}+x}+\lim\limits_{x\rightarrow+\infty}\dfrac{x^2-x-x^2}{\sqrt{x^2-x}+x}=\dfrac{\dfrac{1}{x}}{1+1}-\dfrac{\dfrac{x}{x}}{1+1}=-\dfrac{1}{2}\)
28 tháng 2 2021
\(\lim\limits_{x\rightarrow-2}\dfrac{x^3+2x^2}{\sqrt{x^2+4x+4}}=\lim\limits_{x\rightarrow-2}\dfrac{x^2\left(x+2\right)}{\sqrt{\left(x+2\right)^2}}\)
\(=\lim\limits_{x\rightarrow-2}x^2=\left(-2\right)^2=4\)
p/s: bài này mình chưa học trên lớp nên ko chắc 100% đúng
NV
Nguyễn Việt Lâm
Giáo viên
28 tháng 2 2021
\(\lim\limits_{x\rightarrow+\infty}\dfrac{\sqrt{x+1}}{\sqrt{x+\sqrt{x+1}}+\sqrt{x}}=\lim\limits_{x\rightarrow+\infty}\dfrac{\sqrt{1+\dfrac{1}{x}}}{\sqrt{1+\sqrt{\dfrac{1}{x}+\dfrac{1}{x^2}}}+1}=\dfrac{1}{1+1}=\dfrac{1}{2}\)
Câu c số 1 trong hay ngoài căn nhỉ?
HT
9 tháng 2 2021
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/ \(=\lim\limits_{x\rightarrow+\infty}x.\left(\dfrac{x^2+2x-x^2}{\sqrt{x^2+2x}+x}+2.\dfrac{x^2-x^2+x}{\sqrt{x^2-x}+x}\right)=+\infty\)
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/ \(\lim\limits_{x\rightarrow\pm\infty}\dfrac{\dfrac{2x^2}{x^2}-\dfrac{1}{x^2}}{\dfrac{3}{x^2}-\dfrac{x^2}{x^2}}=\dfrac{2}{-1}=-2\)