HOC24
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Chủ đề / Chương
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\(x\rightarrow-\infty\Rightarrow\left|x\right|=-x\)
\(\Rightarrow\lim\limits_{x\rightarrow-\infty}\dfrac{\sqrt{x^2-x+3}}{2\left|x\right|}=\lim\limits_{x\rightarrow-\infty}\dfrac{-\sqrt{x^2-x+3}}{2x}=\lim\limits_{x\rightarrow-\infty}\dfrac{\sqrt{\dfrac{x^2}{x^2}-\dfrac{x}{x^2}+\dfrac{3}{x^2}}}{\dfrac{2x}{x}}=\dfrac{1}{2}\)
\(=\lim\limits_{x\rightarrow+\infty}\dfrac{3x}{x^2+2}-\lim\limits_{x\rightarrow+\infty}\dfrac{5\sin2x}{x^2+2}+\lim\limits_{x\rightarrow+\infty}\dfrac{\cos^2x}{x^2+2}\)
\(\lim\limits_{x\rightarrow+\infty}\dfrac{3x}{x^2+2}=\lim\limits_{x\rightarrow+\infty}\dfrac{\dfrac{3x}{x^2}}{\dfrac{x^2}{x^2}+\dfrac{2}{x^2}}=0\)
\(-1\le\sin2x\le1\Rightarrow\dfrac{-5}{x^2+2}\le\dfrac{5\sin2x}{x^2+2}\le\dfrac{5}{x^2+2}\)
\(\lim\limits_{x\rightarrow+\infty}-\dfrac{5}{x^2+2}=\lim\limits_{x\rightarrow+\infty}\dfrac{5}{x^2+2}=0\Rightarrow\lim\limits_{x\rightarrow+\infty}\dfrac{5\sin2x}{x^2+2}=0\)
\(0\le\cos^2x\le1\Rightarrow0\le\dfrac{\cos^2x}{x^2+2}\le\dfrac{1}{x^2+2}\)
\(\lim\limits_{x\rightarrow+\infty}\dfrac{1}{x^2+2}=0\Rightarrow\lim\limits_{x\rightarrow+\infty}\dfrac{\cos^2x}{x^2+2}=0\)
\(\Rightarrow\lim\limits_{x\rightarrow+\infty}\dfrac{3x-5\sin2x+\cos^2x}{x^2+2}=0\)
\(=\lim\limits_{x\rightarrow+\infty}\dfrac{16x^4+3x+1-16x^4}{\sqrt[4]{\left(16x^4+3x+1\right)^3}+2x\sqrt[3]{\left(16x^4+3x+1\right)^2}+4x^2\sqrt[3]{16x^4+3x+1}+8x^3}+\lim\limits_{x\rightarrow+\infty}\dfrac{4x^2-4x^2-2}{2x+\sqrt{4x^2+2}}\)
\(=\lim\limits_{x\rightarrow+\infty}\dfrac{\dfrac{3x}{x^3}+\dfrac{1}{x^3}}{\dfrac{\sqrt[4]{\left(16x^4+3x+1\right)^3}}{x^3}+\dfrac{2x\sqrt[3]{\left(16x^4+3x+1\right)^2}}{x^3}+\dfrac{4x^2\sqrt[3]{16x^4+3x+1}}{x^3}+\dfrac{8x^3}{x^3}}+\lim\limits_{x\rightarrow+\infty}\dfrac{-\dfrac{2}{x}}{\dfrac{2x}{x}+\dfrac{\sqrt{4x^2+2}}{x}}=0\)
a/ \(=\lim\limits_{x\rightarrow-\infty}\dfrac{4x^2-4x^2+x-1}{2x-\sqrt{4x^2-x+1}}=\lim\limits_{x\rightarrow-\infty}\dfrac{\dfrac{x}{x}-\dfrac{1}{x}}{\dfrac{2x}{x}+\sqrt{\dfrac{4x^2}{x^2}-\dfrac{x}{x^2}+\dfrac{1}{x^2}}}=\dfrac{1}{2+2}=\dfrac{1}{4}\)
b/ \(=\lim\limits_{x\rightarrow-\infty}x^2\left(-\sqrt{\dfrac{4x^2}{x^2}+\dfrac{1}{x^2}}-\dfrac{x}{x}\right)=\lim\limits_{x\rightarrow-\infty}x^2.\left(-3\right)=-\infty\)
a/ \(=\lim\limits_{x\rightarrow-\infty}x^2\left(1+\dfrac{x}{x^2}-\dfrac{1}{x^2}\right)=+\infty\)
b/ \(=\lim\limits_{x\rightarrow+\infty}\dfrac{x^2+x+1-x^2}{\sqrt{x^2+x+1}+x}+\lim\limits_{x\rightarrow+\infty}2.\dfrac{x^2-x^2+x}{\sqrt{x^2-x}+x}\)
\(=\lim\limits_{x\rightarrow+\infty}\dfrac{\dfrac{x}{x}+\dfrac{1}{x}}{\sqrt{\dfrac{x^2}{x^2}+\dfrac{x}{x^2}+\dfrac{1}{x^2}}+\dfrac{x}{x}}+2\lim\limits_{x\rightarrow+\infty}\dfrac{\dfrac{x}{x}}{\sqrt{\dfrac{x^2}{x^2}-\dfrac{x}{x^2}}+\dfrac{x}{x}}=\dfrac{1}{2}+\dfrac{2}{2}=\dfrac{3}{2}\)
c/ \(=\lim\limits_{x\rightarrow+\infty}x\left(\dfrac{x^2+2x-x^2}{\sqrt{x^2+2x}+x}+2.\dfrac{x^2-x^2-x}{\sqrt{x^2+x}+x}\right)\)
\(=\lim\limits_{x\rightarrow+\infty}\dfrac{\dfrac{2x^2}{x}}{\sqrt{\dfrac{x^2}{x^2}+\dfrac{2x}{x^2}+\dfrac{x}{x^2}}}+2\lim\limits_{x\rightarrow+\infty}\dfrac{-\dfrac{x^2}{x}}{\sqrt{\dfrac{x^2}{x^2}+\dfrac{x}{x^2}}+\dfrac{x}{x}}=0\)
a/ \(=\lim\limits_{x\rightarrow-\infty}\dfrac{\dfrac{x\sqrt{x^2+1}}{x}-\dfrac{2x}{x}+\dfrac{1}{x}}{\sqrt[3]{\dfrac{2x^3}{x^3}-\dfrac{2x}{x^3}}+\dfrac{1}{x}}=0\)
b/ \(=\lim\limits_{x\rightarrow-\infty}\dfrac{\dfrac{8x^7}{x^7}}{\dfrac{\left(-2x^7\right)}{x^7}}=-\dfrac{8}{2^7}\)
c/ \(=\lim\limits_{x\rightarrow+\infty}\dfrac{\sqrt{\dfrac{4x^2}{x^2}+\dfrac{x}{x^2}}+\sqrt[3]{\dfrac{8x^3}{x^3}+\dfrac{x}{x^3}-\dfrac{1}{x^3}}}{\sqrt[4]{\dfrac{x^4}{x^4}+\dfrac{3}{x^4}}}=\dfrac{2+2}{1}=4\)
\(=\dfrac{4.3-3}{3-3}=\dfrac{9}{0}=+\infty\)
b/ \(=\lim\limits_{x\rightarrow+\infty}\sqrt{\dfrac{\dfrac{x^2}{x^4}+\dfrac{1}{x^4}}{\dfrac{2x^4}{x^4}+\dfrac{x^2}{x^4}-\dfrac{3}{x^4}}}=0\)
\(=\dfrac{1-\sqrt{1+2}}{1-\sqrt[3]{3+2}}=\dfrac{1-\sqrt{3}}{1-\sqrt[3]{5}}\)
\(=\lim\limits_{x\rightarrow+\infty}\dfrac{5}{3.\left(+\infty\right)+2}=0\)