Giải thích các bước giải:
5.$\lim (\sqrt{n^2+3n-1}+n)=+\infty +\infty=+\infty$
$\begin{split}6.\lim (\sqrt{n^2+3n-1}-n)&=\lim\dfrac{n^2+3n-1-n^2}{\sqrt{n^2+3n-1}+n}\\&=\lim\dfrac{3n-1}{\sqrt{n^2+3n-1}+n}\\&=\lim\dfrac{3-\dfrac 1n}{\sqrt{1+\dfrac 3n-\dfrac{1}{n^2}}+1}\\&=\dfrac 32\end{split}$
7.$\lim(\sqrt{n^2+5n}-n)=\lim \dfrac{n^2+5n-n^2}{\sqrt{n^2+5n}+n}=\lim\dfrac{5n}{\sqrt{n^2+5n}+n}=\lim\dfrac{5}{\sqrt{1+\dfrac 5n}+1}=\dfrac 52$