Chương 4: GIỚI HẠN
\(=\lim\limits_{x\rightarrow-\infty}\dfrac{x^2-4x-x^2}{\sqrt{x^2-4x}+x}=\lim\limits_{x\rightarrow-\infty}\dfrac{-4x}{\left|x\right|\sqrt{1-\dfrac{4x}{x^2}}+x}=-\dfrac{4}{-1+1}=-\infty\)
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\(=\lim\limits_{x\rightarrow-\infty}\dfrac{-\dfrac{3x^2}{x}-\dfrac{1}{x}}{\dfrac{2x}{x}+\dfrac{1}{x}}=\lim\limits_{x\rightarrow-\infty}\dfrac{-3x}{2}=\dfrac{+\infty}{2}=+\infty\)
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\(=\lim\limits_{x\rightarrow+\infty}\dfrac{-\dfrac{2x^2}{x}+\dfrac{x}{x}-\dfrac{1}{x}}{\dfrac{x}{x}+\dfrac{9}{x}}=\lim\limits_{x\rightarrow+\infty}\dfrac{-2x+1}{1}=\dfrac{-\infty}{1}=-\infty\)
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\(\lim\limits_{x\rightarrow-\infty}\dfrac{17}{x^2+1}=\dfrac{17}{\infty}=0\).
Không bt ntn có đúng không.
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\(=\lim\limits_{x\rightarrow+\infty}\left(\sqrt{x^4+3x^2}-x^2\right)=\lim\limits_{x\rightarrow+\infty}\dfrac{x^4+3x^2-x^4}{\sqrt{x^4+3x^2}+x^2}=\dfrac{3}{1+1}=\dfrac{3}{2}\)
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\(\lim\limits_{x\rightarrow+\infty}x\left(\sqrt{x^2+3}-x\right)=\lim\limits_{x\rightarrow+\infty}\dfrac{3x}{\sqrt{x^2+3}+x}=\lim\limits_{x\rightarrow+\infty}\dfrac{3}{\sqrt{1+\dfrac{3}{x^2}}+1}=\dfrac{3}{\sqrt{1}+1}=\dfrac{3}{2}\).
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