Question

cho a, b là sô thực biết lim(x-> \(+\infty\)) (\(ax+b-\sqrt{x^2-6x+2}\)) =3 . tinhs tong \(2ab+b+a^2\)

Answer 1

Để giới hạn đã cho là hữu hạn thì \(a=1\)

\(\lim\limits_{x\rightarrow+\infty}\left(x+b-\sqrt{x^2-6x+2}\right)=\lim\limits_{x\rightarrow+\infty}\dfrac{x^2+2bx+b^2-\left(x^2-6x+2\right)}{x+b+\sqrt{x^2-6x+2}}\)

\(=\lim\limits_{x\rightarrow+\infty}\dfrac{\left(2b+6\right)x+b^2-2}{x+b+\sqrt{x^2-6x+2}}=\lim\limits_{x\rightarrow+\infty}\dfrac{2b+6+\dfrac{b^2-2}{x}}{1+\dfrac{b}{x}+\sqrt{1-\dfrac{6}{x}+\dfrac{2}{x^2}}}=\dfrac{2b+6}{2}=b+3\)

\(\Rightarrow b+3=3\Rightarrow b=0\Rightarrow\left\{{}\begin{matrix}a=1\\b=0\end{matrix}\right.\)