Câu 21: \(\lim_{x\to-\infty}\left(\sqrt{x^2+x}+2x\right)\)
\(=\lim_{x\to-\infty}\left(-x\cdot\sqrt{1+\frac{1}{x}}+2x\right)=\lim_{x\to-\infty}\left\lbrack-x\left(\sqrt{1+\frac{1}{x}}-2\right)\right\rbrack\)
=-∞ vì \(\lim_{x\to-\infty}-x=-\left(-\infty\right)=+\infty\) và \(\lim_{x\to-\infty}\sqrt{1+\frac{1}{x}}-2=\sqrt{1+\frac10}-2=\sqrt1-2=-1<0\)
=>Chọn B
Câu 20:
\(\lim_{x\to-\infty}\left(3x-\sqrt{9x^2-1}\right)=\lim_{x\to-\infty}\left(3x-3\sqrt{x^2-\frac19}\right)\)
\(=\lim_{x\to-\infty}\left(3x-3\cdot\left(-x\right)\cdot\sqrt{1-\frac{1}{9x^2}}\right)=\lim_{x\to-\infty}\left\lbrack3x+3x\cdot\sqrt{1-\frac{1}{9x^2}}\right\rbrack\)
\(=\lim_{x\to-\infty}\left\lbrack3x\left(1+\sqrt{1-\frac{1}{9x^2}}\right)\right\rbrack=-\infty\) vì \(\lim_{x\to-\infty}3x=-\infty;\lim_{x\to-\infty}\left(1+\sqrt{1-\frac{1}{9x^2}}\right)=1+1=2>0\)
=>Chọn C
Câu 19: \(\lim_{x\to0}\left(\frac{1}{x}-\frac{1}{x^2}\right)=\lim_{x\to0}\frac{x-1}{x^2}=-\infty\)
vì \(\lim_{x\to0}x-1=0-1=-1;\lim_{x\to0}x^2=0\)
=>Chọn C
Câu 18:
\(\lim_{x\to-\infty}\left(x^3-3x^2+2\right)\)
\(=\lim_{x\to-\infty}\left\lbrack x^3\left(1-\frac{3}{x}+\frac{2}{x^3}\right)\right\rbrack=-\infty\)
vì \(\lim_{x\to-\infty}x^3=-\infty;\lim_{x\to-\infty}1-\frac{3}{x}+\frac{2}{x^3}=1-0+0=1>0\)
=>Chọn B
Câu 16:
\(\lim_{x\to-\infty}\frac{2x-3}{\sqrt{x^2+1}-x}=\lim_{x\to-\infty}\frac{2x-3}{-x\cdot\sqrt{1+\frac{1}{x^2}}-x}\)
\(=\lim_{x\to-\infty}\frac{2-\frac{3}{x}}{-\sqrt{1+\frac{1}{x^2}}-1}=\frac{2}{-1-1}=\frac{2}{-2}=-1\)
=>Chọn D
