2.
\(\lim\limits_{x\rightarrow-\infty}\left(\sqrt[3]{x^3+x}-x\right)\\ =\lim\limits_{x→-\infty}\dfrac{x^3+x-x^3}{\left(\sqrt[3]{x^3+x}\right)^2+x\sqrt[3]{x^3+x}+x^2}\\ =\lim\limits_{x\rightarrow-\infty}\dfrac{\dfrac{1}{x}}{\left(\sqrt[3]{1+\dfrac{1}{x^2}}\right)^2+\sqrt{1+\dfrac{1}{x^2}}+1}\\ =\lim\limits_{x\rightarrow-\infty}\dfrac{0}{1^2+1+1}=0\)
3.
\(\lim\limits_{x\rightarrow-\infty}\left(\sqrt{4x^2+x+3}+2x+1\right)\\ =\lim\limits_{x\rightarrow-\infty}x\left(\sqrt{4+\dfrac{1}{x}+\dfrac{3}{x^2}}+2+\dfrac{1}{x}\right)\)
Mà \(\lim\limits_{x→-\infty}x=-\infty\)
\(\lim\limits_{x\rightarrow-\infty}\left(\sqrt{4+\dfrac{1}{x}+\dfrac{3}{x^2}}+2+\dfrac{1}{x}\right)=\sqrt{4}+2=4\) > 0
Vậy \(\lim\limits_{x\rightarrow-\infty}\left(\sqrt{4x^2+x+3}+2x+1\right)=-\infty\)
4.
\(\lim\limits_{x\rightarrow+\infty}\dfrac{x^4-x^3+11}{2x-7}\\ =\lim\limits_{x\rightarrow+\infty}\dfrac{x^4\left(1-\dfrac{1}{x}+\dfrac{11}{x^4}\right)}{x\left(2-\dfrac{7}{x}\right)}\\ =\lim\limits_{x\rightarrow+\infty}x^3\left(\dfrac{1-\dfrac{1}{x}+\dfrac{11}{x^4}}{2-\dfrac{7}{x}}\right)\)
Mà \(\lim\limits_{x\rightarrow+\infty}x^3=+\infty\)
\(\lim\limits_{x→+\infty}\left(\dfrac{1-\dfrac{1}{x}+\dfrac{11}{x^4}}{2-\dfrac{7}{x}}\right)=\dfrac{1-0+0}{2-0}=\dfrac{1}{2}\)> 0
Vậy \(\lim\limits_{x\rightarrow+\infty}\dfrac{x^4-x^3+11}{2x-7}=+\infty\)
Lời giải:
\(\lim\limits_{x\to -\infty}\frac{|x|+\sqrt{x^2+10}}{2x+3}=\lim\limits_{x\to -\infty}\frac{\frac{|x|}{x}-\sqrt{1+\frac{10}{x^2}}}{2+\frac{3}{x}}\)
\(=\frac{-1-1}{2}=-1\)
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