1.
\(\lim\left(\sqrt{4n^2+2n+1}-\left(an-b\right)\right)=\lim\dfrac{4n^2+2n+1-\left(an-b\right)^2}{\sqrt{4n^2+2n+1}+an-b}\)
\(=\lim\dfrac{\left(4-a^2\right)n^2+\left(2+ab\right)n+1-b^2}{\sqrt{4n^2+2n+1}+an-b}\)
\(=\lim\dfrac{\left(4-a^2\right)n+2+ab+\dfrac{1-b^2}{n}}{\sqrt{4+\dfrac{2}{n}+\dfrac{1}{n^2}}+a-\dfrac{b}{n}}\)
- Nếu \(4-a^2\ne0\Rightarrow\) giới hạn đã cho đạt giá trị dương vô cực \(\Rightarrow\) ktm
\(\Rightarrow4-a^2=0\Rightarrow\left[{}\begin{matrix}a=2\\a=-2\end{matrix}\right.\)
- Với \(a=-2\Rightarrow\lim\dfrac{\left(4-a^2\right)n+2+ab+\dfrac{1-b^2}{n}}{\sqrt{4+\dfrac{2}{n}+\dfrac{1}{n^2}}+a-\dfrac{b}{n}}=-\infty\) (ktm)
- Với \(a=2\Rightarrow\lim\dfrac{\left(4-a^2\right)n+2+ab+\dfrac{1-b^2}{n}}{\sqrt{4+\dfrac{2}{n}+\dfrac{1}{n^2}}+a-\dfrac{b}{n}}=\dfrac{2+2b}{4}\)
\(\Rightarrow\dfrac{b+1}{2}=1\Rightarrow b=1\)
Vậy \(a=2;b=1\)
Câu 2 làm tương tự
