Bài 2: Giới hạn của hàm số

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

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

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

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

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

\(=\left(\dfrac{\dfrac{0}{\sqrt[3]{\left(0^2+1\right)^2}+\sqrt[3]{0^2+1}+1}+\dfrac{2}{\sqrt{1-2\cdot0}+1}}{0+1}\right)\)

\(=\left(\dfrac{2}{1+1}:1\right)=\dfrac{2}{2}=1\)

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

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

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

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

\(=\dfrac{\sqrt[3]{\left(0+1\right)^2}+\sqrt[3]{0+1}+1}{\sqrt{0+1}+1}=\dfrac{1+1+1}{1+1}=\dfrac{3}{2}\)

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

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

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

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

\(=1-\sqrt[3]{2\cdot0-1}+\sqrt[3]{\left(2\cdot0-1\right)^2}\)

\(=1-\sqrt[3]{-1}+\sqrt[3]{\left(-1\right)^2}\)

\(=1-\left(-1\right)+\left(-1\right)=1\)

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

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

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

\(=\lim\limits_{x\rightarrow+\infty}\dfrac{x\sqrt{x}+\sqrt{x}}{x^2\cdot\sqrt{2+\dfrac{1}{x^2}+\dfrac{1}{x^4}}}\)

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

\(\lim\limits_{x\rightarrow-\infty}\dfrac{\sqrt{x^4-x}}{1-2x}\)

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

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

\(=+\infty\) vì \(\left\{{}\begin{matrix}\lim\limits_{x\rightarrow-\infty}x=-\infty\\\lim\limits_{x\rightarrow-\infty}\sqrt{1-\dfrac{1}{x^3}}=1>0\\\lim\limits_{x\rightarrow-\infty}\dfrac{1}{x}-2=0-2=-2< 0\end{matrix}\right.\)

\(\lim\limits_{x\rightarrow-\infty}\dfrac{2\left|x\right|+3}{\sqrt{x^2+x+5}}\)

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

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

\(\lim\limits_{x\rightarrow-\infty}\dfrac{\sqrt{x^4+4}}{x+4}\)

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

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

\(=\lim\limits_{x\rightarrow-\infty}\dfrac{x\cdot\sqrt{1+\dfrac{1}{x^4}}}{1+\dfrac{4}{x}}\)

\(=-\infty\) vì \(\left\{{}\begin{matrix}\lim\limits_{x\rightarrow-\infty}x=-\infty\\\lim\limits_{x\rightarrow-\infty}\dfrac{\sqrt{1+\dfrac{1}{x^4}}}{1+\dfrac{4}{x}}=\dfrac{1}{1}=1>0\end{matrix}\right.\)

\(\lim\limits_{x\rightarrow-\infty}\dfrac{3x^2-x+7}{2x^3-1}\\ =\lim\limits_{x\rightarrow-\infty}\dfrac{\dfrac{3x^2-x+7}{x^3}}{\dfrac{2x^3-1}{x^3}}=\lim\limits_{x\rightarrow-\infty}\dfrac{\dfrac{3}{x}-\dfrac{1}{x^2}+\dfrac{1}{x^3}}{2-\dfrac{1}{x^3}}=\dfrac{0-0+0}{2-0}=0\)

\(\lim\limits_{x\rightarrow1}\dfrac{5}{\left(x-1\right)\left(x^2-3x+2\right)}\)

\(=\lim\limits_{x\rightarrow1}\dfrac{5}{x^3-3x^2+2x-x^2+3x-2}\)

\(=\lim\limits_{x\rightarrow1}\dfrac{5}{x^3-4x^2+5x-2}\)

\(=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{5}{x^3}}{1-\dfrac{4}{x}+\dfrac{5}{x^2}-\dfrac{2}{x^3}}=0\)

\(\left\{{}\begin{matrix}\lim\limits_{x\rightarrow1}5=5>0\\\lim\limits_{x\rightarrow1}\left(x-1\right)\left(x^2-3x+2\right)=0\\\left(x-1\right)\left(x^2-3x+2\right)=\left(x-2\right)\left(x-1\right)^2< 0\end{matrix}\right.\) 

Suy ra: \(\lim\limits_{x\rightarrow1}\dfrac{5}{\left(x-1\right)\left(x^2-3x+2\right)}=-\infty\)

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

\(=-\infty\) vì \(\left\{{}\begin{matrix}\lim\limits_{x\rightarrow1}\left(x-1\right)^2=\left(1-1\right)^2=0\\\lim\limits_{x\rightarrow1}\dfrac{2x+1}{2x-3}=\dfrac{2\cdot1+1}{2\cdot1-3}=\dfrac{3}{-1}=-3< 0\\\lim\limits_{x\rightarrow1}2=2>0\\\end{matrix}\right.\)