Câu 11: \(\lim_{x\to-\infty}\frac{2x^2+5x-3}{x^2+6x+3}=\lim_{x\to-\infty}\frac{2+\frac{5}{x}-\frac{3}{x^2}}{1+\frac{6}{x}+\frac{3}{x^2}}=\frac{2+0-0}{1+0+0}=\frac21=2\)
=>Chọn D
Câu 12: \(\lim_{x\to+\infty}\frac{x-3}{6x^2+x+1}=\lim_{x\to+\infty}\frac{\frac{1}{x}-\frac{3}{x^2}}{6+\frac{1}{x}+\frac{1}{x^2}}=0\)
=>Chọn D
Câu 13: \(\lim_{x\to-\infty}\frac{2x^3+5x^2-3}{x^2+6x+3}=\lim_{x\to-\infty}\frac{x^3\left(2+\frac{5}{x}-\frac{3}{x^3}\right)}{x^2\left(1+\frac{6}{x}+\frac{3}{x^2}\right)}\)
\(=\lim_{x\to-\infty}\frac{x\cdot\left(2+\frac{5}{x}-\frac{3}{x^3}\right)}{1+\frac{6}{x}+\frac{3}{x^2}}=-\infty\)
=>Chọn C
Câu 14:
\(\lim_{x\to+\infty}\frac{\sqrt{4x^2-2x+1}+2-x}{\sqrt{9x^2-3x}+2x}=\lim_{x\to+\infty}\frac{\sqrt{4-\frac{2}{x}+\frac{1}{x^2}}+\frac{2}{x}-1}{\sqrt{9-\frac{3}{x}}+2}\)
\(=\frac{2+0-1}{3+2}=\frac15\)
=>Chọn D
Bài 1:
a: Vì \(2x^3-3x^2+4x+1\) có bậc là 3
và \(x^4-5x^3+2x^2-x+3\) có bậc là 4
và 3<4
nên \(A=\lim_{x\to+\infty}\frac{2x^3-3x^2+4x+1}{x^4-5x^3+2x^2-x+3}=0\)
b: \(B=\lim_{x\to-\infty}\frac{x+\sqrt{x^2+2}}{\sqrt[3]{8x^3+x^2+1}}=\lim_{x\to-\infty}\frac{x-x\cdot\sqrt{1+\frac{2}{x^2}}}{x\cdot\sqrt[3]{8+\frac{1}{x}+\frac{1}{x^3}}}\)
\(=\lim_{x\to-\infty}\frac{1-\sqrt{1+\frac{2}{x^2}}}{\sqrt[3]{8+\frac{1}{x}+\frac{1}{x^3}}}=\frac{1-1}{\sqrt[3]{8+0+0}}=0\)
e: \(E=\lim_{x\to+\infty}\frac{3x^2-x+7}{2x^3-1}\)
\(=\lim_{x\to+\infty}\frac{x^2\left(3-\frac{1}{x}+\frac{7}{x^2}\right)}{x^3\left(2-\frac{1}{x^3}\right)}=\lim_{x\to+\infty}\frac{3-\frac{1}{x}+\frac{7}{x^2}}{x\left(2-\frac{1}{x^3}\right)}\)
\(=+\infty\) vì \(\lim_{x\to+\infty}\frac{3-\frac{1}{x}+\frac{7}{x^2}}{2-\frac{1}{x^3}}=\frac{3-0+0}{2-0}=\frac32>0\) và \(\lim_{x\to+\infty}\frac{1}{x}=+\infty\)
f: \(F=\lim_{x\to-\infty}\frac{2x^3+3x-4}{-x^3-x^2+1}\)
\(=\lim_{x\to-\infty}\frac{2+\frac{3}{x^2}-\frac{4}{x^3}}{-1-\frac{1}{x}+\frac{1}{x^3}}=\frac{2}{-1}=-2\)
g: \(G=\lim_{x\to+\infty}\frac{3x^2-x+3}{x-4}\)
\(=\lim_{x\to+\infty}\frac{x^2\left(1-\frac{1}{x}+\frac{3}{x^2}\right)}{x\left(1-\frac{4}{x}\right)}=\lim_{x\to+\infty}\left\lbrack x\cdot\frac{1-\frac{1}{x}+\frac{3}{x^2}}{1-\frac{4}{x}}\right\rbrack=+\infty\)
h: \(H=\lim_{x\to-\infty}\frac{2x^2-2x+3}{-x+5}\)
\(=\lim_{x\to-\infty}\frac{x^2\left(2-\frac{2}{x}+\frac{3}{x^2}\right)}{x\left(-1+\frac{5}{x}\right)}=\lim_{x\to-\infty}\left\lbrack x\cdot\frac{2-\frac{2}{x}+\frac{3}{x^2}}{-1+\frac{5}{x}}\right\rbrack=+\infty\)
vì \(\lim_{x\to-\infty}x=-\infty\) và \(\lim_{x\to-\infty}\frac{2-\frac{2}{x}+\frac{3}{x^2}}{-1+\frac{5}{x}}=\frac{2-0+0}{-1+0}=\frac{2}{-1}=-2<0\)
