1) Ta có
$\lim \dfrac{\sqrt{n^2 + 1}}{2n + 3} = \lim \dfrac{\sqrt{1 + \frac{1}{n^2}}}{2 + \frac{3}{n}} = \dfrac{1}{2}$
2) Ta có
$\lim \dfrac{\sqrt[3]{n^3 + 1} - 1}{\sqrt{n^2 + 3} - 2} = \lim \dfrac{\sqrt[3]{1 + \frac{1}{n^3}} - \frac{1}{n}}{\sqrt{1 + \frac{3}{n^2}} - \frac{2}{n}} = \dfrac{1 - 0}{1 - 2} = -1$.