giúp e vs ạ :((
b.
Với \(a>0\Rightarrow\lim\limits_{x\rightarrow+\infty}x\left(\sqrt{4+\dfrac{2}{x}+\dfrac{1}{x^2}}+a+\dfrac{b}{x}\right)=+\infty\) (ktm)
Với \(a< 0\):
\(\lim\limits_{x\rightarrow+\infty}\left(\sqrt{4x^2+2x+1}+ax+b\right)=\lim\limits_{x\rightarrow+\infty}\dfrac{4x^2+2x+1-a^2x^2-2abx-b^2}{\sqrt{4x^2+2x+1}-ax-b}\)
\(=\lim\limits_{x\rightarrow+\infty}\dfrac{\left(4-a^2\right)x^2+\left(2-2ab\right)x+1-b^2}{\sqrt{4x^2+2x+1}-ax-b}=\lim\limits_{x\rightarrow+\infty}\dfrac{\left(4-a^2\right)x+\left(2-2ab\right)+\dfrac{1-b^2}{x}}{\sqrt{4+\dfrac{2}{x}+\dfrac{1}{x^2}}-a-\dfrac{b}{x}}\) (1)
Giới hạn đã cho hữu hạn khi \(4-a^2=0\Rightarrow a=\pm2\Rightarrow a=-2\)
Thế vào (1):
\(\lim\limits_{x\rightarrow+\infty}\dfrac{\left(2+4b\right)+\dfrac{1-b^2}{x}}{\sqrt{4+\dfrac{2}{x}+\dfrac{1}{x^2}}+2-\dfrac{b}{x}}=\dfrac{2+4b}{2+2}=\dfrac{2b+1}{2}=\dfrac{2}{3}\)
\(\Rightarrow b=\dfrac{1}{6}\)
a.
\(\lim\limits_{x\rightarrow1}\dfrac{2x^2+ax+b}{\left(x-1\right)\left(x+1\right)}\) hữu hạn khi \(2x^2+ax+b=0\) có nghiệm \(x=1\)
\(\Rightarrow2.1+a.1+b=0\Rightarrow b=-a-2\)
Thế vào: \(\dfrac{1}{4}=\lim\limits_{x\rightarrow1}\dfrac{2x^2+ax-a-2}{\left(x-1\right)\left(x+1\right)}=\lim\limits_{x\rightarrow1}\dfrac{\left(x-1\right)\left(2x+2+a\right)}{\left(x-1\right)\left(x+1\right)}=\lim\limits_{x\rightarrow1}\dfrac{2x+2+a}{x+1}=\dfrac{a+4}{2}\)
\(\Rightarrow a+4=\dfrac{1}{2}\Rightarrow a=-\dfrac{7}{2}\Rightarrow b=-a-2=\dfrac{3}{2}\)
