`~rai~`
$\begin{array}{I}a)lim\sqrt{n^2+2n}-\sqrt{n^2+1}\\=lim\dfrac{\left(\sqrt{n^2+2n}-\sqrt{n^2+1}\right)\left(\sqrt{n^2+2n}+\sqrt{n^2+2n}\right)}{\sqrt{n^2+2n}+\sqrt{n^2+1}}\\=lim\dfrac{n^2+2n-n^2-1}{\sqrt{n^2+2n}+\sqrt{n^2+1}}\\lim\dfrac{2n-1}{\sqrt{n^2+2n}+\sqrt{n^2+1}}\\=lim\dfrac{2+\dfrac{1}{n}}{\sqrt{1+\dfrac{2}{n}}+\sqrt{1+\dfrac{1}{n^2}}}\\=\dfrac{2}{\sqrt{1}+\sqrt{1}}\\=1.\\b)C1:lim\dfrac{2n^3+3}{1+2n^2}\\=lim\dfrac{2+\dfrac{3}{n^3}}{\dfrac{1}{n^3}+\dfrac{2}{n}}\\=+\infty.\\C2:lim\dfrac{2n^3+3}{1+2n^2}\\=lim\dfrac{n^3\left(2+\dfrac{3}{n^3}\right)}{n^2\left(\dfrac{1}{n^2}+2\right)}\\=lim n.\dfrac{2+\dfrac{3}{n^3}}{\dfrac{1}{n^2}+2}\\\text{Ta có:}lim n=+\infty;lim\dfrac{2+\dfrac{3}{n^3}}{\dfrac{1}{n^2}+2}=\dfrac{2+0}{0+2}=1>0\\\Rightarrow lim n.\dfrac{2+\dfrac{3}{n^3}}{\dfrac{1}{n^2}+2}=+\infty\Rightarrow lim\dfrac{2n^3+3}{1+2n^2}=+\infty.\end{array}$
