\(\lim\left(\sqrt{an^2+3n+1}-2n\right)=\lim\dfrac{an^2+3n+1-4n^2}{\sqrt{an^2+3n+1}+2n}\)
\(=\lim\dfrac{\left(a-4\right)n^2+3n+1}{\sqrt{an^2+3n+1}+2n}=\lim\dfrac{\left(a-4\right)n+3+\dfrac{1}{n}}{\sqrt{a+\dfrac{3}{n}+\dfrac{1}{n^2}}+2}\)
Nếu \(a\ne4\Rightarrow\left\{{}\begin{matrix}\lim\left(a-4\right)n+3+\dfrac{1}{n}=\infty\\\lim\left(\sqrt{a+\dfrac{3}{n}+\dfrac{1}{n^2}}+2\right)=\sqrt{a}+2\end{matrix}\right.\)
\(\Rightarrow\lim\dfrac{\left(a-4\right)n+3+\dfrac{1}{n^2}}{\sqrt{a+\dfrac{3}{n}+\dfrac{1}{n^2}}+2}=\infty\) (ktm)
\(\Rightarrow a=4\in\left(2;+\infty\right)\)
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