Câu 12
Ta có
$\lim \dfrac{\sqrt{4n^2 - 7n + 1} + 3n}{9-5n} = \lim \dfrac{\sqrt{4 - \frac{7}{n} + \frac{1}{n^2}} + 3}{\frac{9}{n} - 5} = \dfrac{2+3}{-5} = -1$
Câu 13
$\lim \dfrac{\sqrt{4n^2 + 5} - \sqrt{n+4}}{2n-1} = \lim \dfrac{\sqrt{4 + \frac{5}{n^2}} - \sqrt{\frac{1}{n} + \frac{4}{n^2}}}{2 - \frac{1}{n}} = \dfrac{2-0}{2} = 1$
