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\(\lim\limits_{x\rightarrow-\infty}f\left(x\right)\)

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

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

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

30 tháng 1 2021

\(\lim\limits_{x\rightarrow1}\dfrac{\sqrt[3]{7+x^3}-\sqrt{3+x^2}}{x-1}=\lim\limits_{x\rightarrow1}\dfrac{\left(\sqrt[3]{7+x^3}-2\right)-\left(\sqrt{3+x^2}-2\right)}{x-1}=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{x^3-1}{\left(\sqrt[3]{7+x^3}\right)^2+2\sqrt[3]{7+x^3}+4}-\dfrac{x^2-1}{\sqrt{3+x^2}+2}}{x-1}=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{x^2+x+1}{\left(\sqrt[3]{7+x^3}\right)^2+2\sqrt[3]{7+x^3}+4}-\dfrac{x+1}{\sqrt{3+x^2}+2}}{1}=\dfrac{3}{12}-\dfrac{2}{4}=\dfrac{1}{4}-\dfrac{1}{2}=-\dfrac{1}{4}\).

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

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

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

\(=\dfrac{-\sqrt{1-0}}{1+0}=-\dfrac{1}{1}=-1\)

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

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

\(=\dfrac{3\sqrt{2}}{2}\)

26 tháng 12 2023

\(\lim\limits_{x\rightarrow1}\dfrac{f\left(x\right)-2x+1}{x-1}=3\rightarrow\lim\limits_{x\rightarrow1}\left(f\left(x\right)-2x+1\right)=0\\ \rightarrow\lim\limits_{x\rightarrow1}f\left(x\right)=1\)

\(\lim\limits_{x\rightarrow1}\dfrac{\sqrt{3f\left(x\right)+1}-x-1}{\sqrt{4x+5}-3x-2}=\dfrac{\sqrt{3.1+1}-1-1}{\sqrt{4.1+5}-3.1-2}=0\)

NV
7 tháng 2 2021

\(a=\lim\limits_{x\rightarrow3}\dfrac{2x+3-x^2}{\left(x^2-4x+3\right)\left(\sqrt[]{2x+3}+x\right)}=\lim\limits_{x\rightarrow3}\dfrac{\left(x-3\right)\left(-x-1\right)}{\left(x-3\right)\left(x-1\right)\left(\sqrt[]{2x+3}+x\right)}\)

\(=\lim\limits_{x\rightarrow3}\dfrac{-x-1}{\left(x-1\right)\left(\sqrt[]{2x+3}+x\right)}=...\)

\(b=\lim\limits_{x\rightarrow0}\dfrac{\left(x+1\right)^{\dfrac{1}{3}}-1}{\left(2x+1\right)^{\dfrac{1}{4}}-1}=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{1}{3}\left(x+1\right)^{-\dfrac{2}{3}}}{\dfrac{1}{2}\left(2x+1\right)^{-\dfrac{3}{4}}}=\dfrac{2}{3}\)

\(c=\lim\limits_{x\rightarrow0}\dfrac{\left(\sqrt[]{1+4x}-2x-1\right)+\left(2x+1-\sqrt[3]{1+6x}\right)}{x^2}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{-4x^2}{2x+1+\sqrt[]{4x+1}}+\dfrac{x^2\left(8x+12\right)}{\left(2x+1\right)^2+\left(2x+1\right)\sqrt[3]{1+6x}+\sqrt[3]{\left(1+6x\right)^2}}}{x^2}\)

\(=\lim\limits_{x\rightarrow0}\left(\dfrac{-4}{2x+1+\sqrt[]{4x+1}}+\dfrac{8x+12}{\left(2x+1\right)^2+\left(2x+1\right)\sqrt[3]{1+6x}+\sqrt[3]{\left(1+6x\right)^2}}\right)=...\)

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/ \(\lim\limits_{x\rightarrow+\infty}\dfrac{x\sqrt{\dfrac{4x^2}{x^2}+\dfrac{x}{x^2}}+x\sqrt[3]{\dfrac{8x^3}{x^3}+\dfrac{x}{x^3}-\dfrac{1}{x^3}}}{x\sqrt[4]{\dfrac{x^4}{x^4}+\dfrac{3}{x^4}}}=\dfrac{2+2}{1}=4\)

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/ \(\lim\limits_{x\rightarrow\pm\infty}\dfrac{\dfrac{2x^2}{x^2}-\dfrac{1}{x^2}}{\dfrac{3}{x^2}-\dfrac{x^2}{x^2}}=\dfrac{2}{-1}=-2\)

NV
7 tháng 2 2021

\(a=\lim\limits_{x\rightarrow0}\dfrac{\sqrt{4x+1}-1+1-\sqrt[3]{2x+1}}{x}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{4x}{\sqrt[]{4x+1}+1}+\dfrac{-2x}{1+\sqrt[3]{2x+1}+\sqrt[3]{\left(2x+1\right)^2}}}{x}\)

\(=\lim\limits_{x\rightarrow0}\left(\dfrac{4}{\sqrt[]{4x+1}+1}+\dfrac{-2}{1+\sqrt[3]{2x+1}+\sqrt[3]{\left(2x+1\right)^2}}\right)=...\)

\(b=\lim\limits_{x\rightarrow1}\dfrac{4\left(x-1\right)\left(\sqrt[3]{\left(5x+3\right)^2}+2\sqrt[3]{5x+3}+4\right)}{5\left(x-1\right)\left(\sqrt[]{4x+5}+3\right)}\)

\(=\lim\limits_{x\rightarrow1}\dfrac{4\left(\sqrt[3]{\left(5x+3\right)^2}+2\sqrt[3]{5x+3}+4\right)}{5\left(\sqrt[]{4x+5}+3\right)}=...\)

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

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