\(\lim\limits_{x\rightarrow4}\dfrac{2x-8}{4x^2-18x+8}=\lim\limits_{x\rightarrow4}\dfrac{2\left(x-4\right)}{2\left(x-4\right)\left(2x-1\right)}=\lim\limits_{x\rightarrow4}\dfrac{1}{2x-1}=-\dfrac{1}{7}\) b. \(\lim\limits_{x\rightarrow-\infty}\left(\sqrt{x^2+3x-1}+x\right)=\lim\limits_{x\rightarrow-\infty}\dfrac{\left(\sqrt{x^2+3x-1}+x\right)\left(\sqrt{x^2+3x-1}-x\right)}{\sqrt{x^2+3x-1}-x}\) \(=\lim\limits_{x\rightarrow-\infty}\dfrac{3x-1}{\sqrt{x^2+3x-1}-x}=\lim\limits_{x\rightarrow-\infty}\dfrac{3-\dfrac{1}{x}}{-\sqrt{1+\dfrac{3}{x}-\dfrac{1}{x^2}}-1}=\dfrac{3-0}{-\sqrt{1+0-0}-1}=-\dfrac{3}{2}\) c. \(\lim\limits_{x\rightarrow1^+}\dfrac{2x^2-3x+1}{x^2-1}=\lim\limits_{x\rightarrow1^+}\dfrac{\left(x-1\right)\left(2x-1\right)}{\left(x-1\right)\left(x+1\right)}=\lim\limits_{x\rightarrow1^+}\dfrac{2x-1}{x+1}=\dfrac{1}{2}\) d. \(\lim\limits_{x\rightarrow-\infty}\dfrac{2x^2+3x}{\sqrt{4x^4+2x^2}+3x^2-1}=\lim\limits_{x\rightarrow-\infty}\dfrac{x^2\left(2+\dfrac{3}{x}\right)}{x^2\left(\sqrt{4+\dfrac{2}{x^2}}+3-\dfrac{1}{x^2}\right)}\) \(=\lim\limits_{x\rightarrow-\infty}\dfrac{2+\dfrac{3}{x}}{\sqrt{4+\dfrac{2}{x^2}}+3-\dfrac{1}{x^2}}=\dfrac{2+0}{\sqrt{4+0}+3-0}=\dfrac{2}{5}\) Câu trả lời: \(\lim\limits_{x\rightarrow4}\dfrac{2x-8}{4x^2-18x+8}=\lim\limits_{x\rightarrow4}\dfrac{2\left(x-4\right)}{2\left(x-4\right)\left(2x-1\right)}=\lim\limits_{x\rightarrow4}\dfrac{1}{2x-1}=-\dfrac{1}{7}\) b. \(\lim\limits_{x\rightarrow-\infty}\left(\sqrt{x^2+3x-1}+x\right)=\lim\limits_{x\rightarrow-\infty}\dfrac{\left(\sqrt{x^2+3x-1}+x\right)\left(\sqrt{x^2+3x-1}-x\right)}{\sqrt{x^2+3x-1}-x}\) \(=\lim\limits_{x\rightarrow-\infty}\dfrac{3x-1}{\sqrt{x^2+3x-1}-x}=\lim\limits_{x\rightarrow-\infty}\dfrac{3-\dfrac{1}{x}}{-\sqrt{1+\dfrac{3}{x}-\dfrac{1}{x^2}}-1}=\dfrac{3-0}{-\sqrt{1+0-0}-1}=-\dfrac{3}{2}\) c. \(\lim\limits_{x\rightarrow1^+}\dfrac{2x^2-3x+1}{x^2-1}=\lim\limits_{x\rightarrow1^+}\dfrac{\left(x-1\right)\left(2x-1\right)}{\left(x-1\right)\left(x+1\right)}=\lim\limits_{x\rightarrow1^+}\dfrac{2x-1}{x+1}=\dfrac{1}{2}\) d. \(\lim\limits_{x\rightarrow-\infty}\dfrac{2x^2+3x}{\sqrt{4x^4+2x^2}+3x^2-1}=\lim\limits_{x\rightarrow-\infty}\dfrac{x^2\left(2+\dfrac{3}{x}\right)}{x^2\left(\sqrt{4+\dfrac{2}{x^2}}+3-\dfrac{1}{x^2}\right)}\) \(=\lim\limits_{x\rightarrow-\infty}\dfrac{2+\dfrac{3}{x}}{\sqrt{4+\dfrac{2}{x^2}}+3-\dfrac{1}{x^2}}=\dfrac{2+0}{\sqrt{4+0}+3-0}=\dfrac{2}{5}\)

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Givemesome Flan

11 tháng 3 2023

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Nguyễn Việt Lâm

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11 tháng 3 2023

\(\lim\limits_{x\rightarrow4}\dfrac{2x-8}{4x^2-18x+8}=\lim\limits_{x\rightarrow4}\dfrac{2\left(x-4\right)}{2\left(x-4\right)\left(2x-1\right)}=\lim\limits_{x\rightarrow4}\dfrac{1}{2x-1}=-\dfrac{1}{7}\)

\(\lim\limits_{x\rightarrow-\infty}\left(\sqrt{x^2+3x-1}+x\right)=\lim\limits_{x\rightarrow-\infty}\dfrac{\left(\sqrt{x^2+3x-1}+x\right)\left(\sqrt{x^2+3x-1}-x\right)}{\sqrt{x^2+3x-1}-x}\)

\(=\lim\limits_{x\rightarrow-\infty}\dfrac{3x-1}{\sqrt{x^2+3x-1}-x}=\lim\limits_{x\rightarrow-\infty}\dfrac{3-\dfrac{1}{x}}{-\sqrt{1+\dfrac{3}{x}-\dfrac{1}{x^2}}-1}=\dfrac{3-0}{-\sqrt{1+0-0}-1}=-\dfrac{3}{2}\)

\(\lim\limits_{x\rightarrow1^+}\dfrac{2x^2-3x+1}{x^2-1}=\lim\limits_{x\rightarrow1^+}\dfrac{\left(x-1\right)\left(2x-1\right)}{\left(x-1\right)\left(x+1\right)}=\lim\limits_{x\rightarrow1^+}\dfrac{2x-1}{x+1}=\dfrac{1}{2}\)

\(\lim\limits_{x\rightarrow-\infty}\dfrac{2x^2+3x}{\sqrt{4x^4+2x^2}+3x^2-1}=\lim\limits_{x\rightarrow-\infty}\dfrac{x^2\left(2+\dfrac{3}{x}\right)}{x^2\left(\sqrt{4+\dfrac{2}{x^2}}+3-\dfrac{1}{x^2}\right)}\)

\(=\lim\limits_{x\rightarrow-\infty}\dfrac{2+\dfrac{3}{x}}{\sqrt{4+\dfrac{2}{x^2}}+3-\dfrac{1}{x^2}}=\dfrac{2+0}{\sqrt{4+0}+3-0}=\dfrac{2}{5}\)

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