Giới hạn lim(căn bậc hai n^2-3n-5 - căn bậc hai 9n^2+3/2n-1) bằng? A.5/2 B.-5/2 C.1 D.-1

Cách 1:

$lim\frac{\sqrt{n^2-3n-5}-\sqrt{9n^2+3}}{2n-1}$

$=lim\frac{\left(\sqrt{n^2-3n-5}-\sqrt{9n^2+3}\right).\left(\sqrt{n^2-3n-5}+\sqrt{9n^2+3}\right)}{\left(\sqrt{n^2-3n-5}+\sqrt{9n^2+3}\right).\left(2n-1\right)}$

$=lim\frac{(n^2-3n-5)-(9n^2+3)}{\left(\sqrt{n^2-3n-5}+\sqrt{9n^2+3}\right).\left(2n-1\right)}$

$=lim\frac{-8n^2-3n-8}{\left(\sqrt{n^2-3n-5}+\sqrt{9n^2+3}\right).\left(2n-1\right)}$

$=lim\frac{-8-\frac{3}{n}-\frac{8}{n^2}}{\left(\sqrt{1-\frac{3}{n}-\frac{5}{n^2}}+\sqrt{9+\frac{3}{n^2}}\right)\left(2-\frac{1}{n}\right)}=\frac{-8}{4.2}=-1$

Cách 2: Chia cả tử và mẫu cho n.

$\lim \frac{\sqrt{n^2-3n-5}-\sqrt{9n^2+3}}{2n-1}=\lim \frac{\sqrt{1-\frac{3}{n}-\frac{5}{n^2}}-\sqrt{9+\frac{3}{n^2}}}{2-\frac{1}{n}}=\lim \frac{1-3}{2}=-1$

Đáp án cần chọn là: D