1a.
\(\lim\limits_{x\rightarrow-\infty}\left(\sqrt{4x^2-x+2}+2x\right)=\lim\limits_{x\rightarrow-\infty}\dfrac{-x+2}{\sqrt{4x^2-x+2}-2x}\)
\(=\lim\limits_{x\rightarrow-\infty}\dfrac{x\left(-1+\dfrac{2}{x}\right)}{\left|x\right|\sqrt{4-\dfrac{1}{x}+\dfrac{2}{x^2}}-2x}=\lim\limits_{x\rightarrow-\infty}\dfrac{x\left(-1+\dfrac{2}{x}\right)}{x\left(-\sqrt{4-\dfrac{1}{x}+\dfrac{2}{x^2}}-2\right)}\)
\(=\dfrac{-1+0}{-\sqrt{4-0+0}-2}=\dfrac{1}{4}\)
b.
\(\lim\limits_{x\rightarrow+\infty}\left(\sqrt{x^2+2x-3}-x\right)=\lim\limits_{x\rightarrow+\infty}\dfrac{2x-3}{\sqrt{x^2+2x-3}+x}\)
\(=\lim\limits_{x\rightarrow+\infty}\dfrac{x\left(2-\dfrac{3}{x}\right)}{x\left(\sqrt{1+\dfrac{2}{x}-\dfrac{3}{x^2}}+1\right)}=\dfrac{2-0}{\sqrt{1+0-0}+1}=1\)
2a.
\(\lim\limits_{x\rightarrow-\infty}\left(-x^3+x^2-x+1\right)=\lim\limits_{x\rightarrow-\infty}x^3\left(-1+\dfrac{1}{x}-\dfrac{1}{x^2}+\dfrac{1}{x^3}\right)\)
Mà \(\left\{{}\begin{matrix}\lim\limits_{x\rightarrow-\infty}x^3=-\infty\\\lim\limits_{x\rightarrow-\infty}\left(-1+\dfrac{1}{x}-\dfrac{1}{x^2}+\dfrac{1}{x^3}\right)=-1< 0\end{matrix}\right.\)
\(\Rightarrow\lim\limits_{x\rightarrow-\infty}x^3\left(-1+\dfrac{1}{x}-\dfrac{1}{x^2}+\dfrac{1}{x^3}\right)=+\infty\)
b.
\(\lim\limits_{x\rightarrow-\infty}\left(2x-\sqrt{4x^2+2x-1}\right)=\lim\limits_{x\rightarrow-\infty}\left(2x-\left|x\right|\sqrt{4+\dfrac{2}{x}-\dfrac{1}{x^2}}\right)\)
\(=\lim\limits_{x\rightarrow-\infty}\left(2x+x\sqrt{4+\dfrac{2}{x}-\dfrac{1}{x^2}}\right)=\lim\limits_{x\rightarrow-\infty}x\left(2+\sqrt{4+\dfrac{2}{x}-\dfrac{1}{x^2}}\right)\)
Do \(\left\{{}\begin{matrix}\lim\limits_{x\rightarrow-\infty}x=-\infty\\\lim\limits_{x\rightarrow-\infty}\left(2+\sqrt{4+\dfrac{2}{x}-\dfrac{1}{x^2}}\right)=2+2=4>0\end{matrix}\right.\)
\(\Rightarrow\lim\limits_{x\rightarrow-\infty}x\left(2+\sqrt{4+\dfrac{2}{x}-\dfrac{1}{x^2}}\right)=-\infty\)
3a.
\(\lim\limits_{x\rightarrow\infty}\dfrac{2x^2-x+3}{x^2+1}=\lim\limits_{x\rightarrow\infty}\dfrac{x^2\left(2-\dfrac{1}{x}+\dfrac{3}{x^2}\right)}{x^2\left(1+\dfrac{1}{x^2}\right)}=\lim\limits_{x\rightarrow\infty}\dfrac{2-\dfrac{1}{x}+\dfrac{3}{x^2}}{1+\dfrac{1}{x^2}}=\dfrac{2}{1}=2\)
b.
\(\lim\limits_{x\rightarrow+\infty}\dfrac{\sqrt{x^2+1}}{x}=\lim\limits_{x\rightarrow+\infty}\dfrac{x\sqrt{1+\dfrac{1}{x^2}}}{x}=\lim\limits_{x\rightarrow+\infty}\sqrt{1+\dfrac{1}{x^2}}=\sqrt{1+0}=1\)
