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HT
6 tháng 3 2021
a/ \(=\lim\limits_{x\rightarrow\pm\infty}\dfrac{\dfrac{\left(2x\right)^2.\left(4x\right)^3}{x^4}}{\dfrac{\left(3x\right)^2\left(5x^2\right)}{x^4}}=\lim\limits_{x\rightarrow\pm\infty}\dfrac{4^4.x}{45}=\pm\infty\)
b/ \(=\lim\limits_{x\rightarrow\pm\infty}\dfrac{\sqrt[3]{\dfrac{x^3}{x^3}+\dfrac{2x^2}{x^3}+\dfrac{x}{x^3}}}{\dfrac{2x}{x}-\dfrac{2}{x}}=\dfrac{1}{2}\)
c/ \(=\lim\limits_{x\rightarrow\pm\infty}\dfrac{\dfrac{\sqrt[3]{\left(x^3+2x^2\right)^2}}{x^2}+\dfrac{x\sqrt[3]{x^3+2x^2}}{x^2}+\dfrac{x^2}{x^2}}{\dfrac{3x^2}{x^2}-\dfrac{2x}{x^2}}=\dfrac{1+1+1}{3}=1\)
d/ \(=\lim\limits_{x\rightarrow+\infty}\dfrac{\dfrac{\left(-3x\right)^3x^2}{x^5}}{-\dfrac{4x^5}{x^5}}=\dfrac{-27}{-4}=\dfrac{27}{4}\)
e/ \(=\lim\limits_{x\rightarrow+\infty}\dfrac{\dfrac{\left(2x\right)^{20}.\left(3x\right)^{20}}{x^{50}}}{\dfrac{\left(2x\right)^{50}}{x^{50}}}=0\)
g/ \(=\lim\limits_{x\rightarrow+\infty}\dfrac{\dfrac{8x^3.\left(4x^5\right)^9}{x^{47}}}{\dfrac{11x^{47}}{x^{47}}}=+\infty\)
C
1
24 tháng 12 2023
TH1: \(\lim\limits_{x\rightarrow+\infty}\dfrac{x+1}{\sqrt{x^2-x+1}}\)
\(=\lim\limits_{x\rightarrow+\infty}\dfrac{1+\dfrac{1}{x}}{\sqrt{1-\dfrac{1}{x}+\dfrac{1}{x^2}}}\)
\(=\dfrac{1+0}{\sqrt{1-0+0}}=\dfrac{1}{1}=1\)
TH2: \(\lim\limits_{x\rightarrow-\infty}\dfrac{x+1}{\sqrt{x^2-x+1}}\)
\(=\lim\limits_{x\rightarrow-\infty}\dfrac{x+1}{-x\cdot\sqrt{1-\dfrac{1}{x}+\dfrac{1}{x^2}}}\)
\(=\lim\limits_{x\rightarrow-\infty}\dfrac{1+\dfrac{1}{x}}{-\sqrt{1-\dfrac{1}{x}+\dfrac{1}{x^2}}}=\dfrac{1+0}{-\sqrt{1-0+0}}=\dfrac{1}{-1}=-1\)
HT
27 tháng 1 2021
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/ \(\lim\limits_{x\rightarrow\pm\infty}\dfrac{\dfrac{2x^2}{x^2}-\dfrac{1}{x^2}}{\dfrac{3}{x^2}-\dfrac{x^2}{x^2}}=\dfrac{2}{-1}=-2\)
C
1
24 tháng 12 2023
Th1: \(\lim\limits_{x\rightarrow+\infty}\dfrac{x}{\sqrt{x^2+x+1}}\)
\(=\lim\limits_{x\rightarrow+\infty}\dfrac{1}{\sqrt{1+\dfrac{1}{x}+\dfrac{1}{x^2}}}\)
\(=\dfrac{1}{\sqrt{1+0+0}}=\dfrac{1}{1}=1\)
TH2: \(\lim\limits_{x\rightarrow-\infty}\dfrac{x}{\sqrt{x^2+x+1}}\)
\(=\lim\limits_{x\rightarrow-\infty}\dfrac{x}{-x\sqrt{1+\dfrac{1}{x}+\dfrac{1}{x^2}}}\)
\(=\lim\limits_{x\rightarrow-\infty}\dfrac{-1}{\sqrt{1+\dfrac{1}{x}+\dfrac{1}{x^2}}}=\dfrac{-1}{1}=-1\)
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/ \(=\lim\limits_{x\rightarrow+\infty}\dfrac{x^2+1-x^2}{\sqrt{x^2+1}+x}+\lim\limits_{x\rightarrow+\infty}\dfrac{x^2-x-x^2}{\sqrt{x^2-x}+x}=\dfrac{\dfrac{1}{x}}{1+1}-\dfrac{\dfrac{x}{x}}{1+1}=-\dfrac{1}{2}\)
R
0
\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)