Giới hạn $\lim \dfrac{{\sqrt {{n^2} - 3n - 5}  - \sqrt {9{n^2} + 3} }}{{2n - 1}}$bằng?

Cách 1:

$\begin{array}{l}\lim \frac{{\sqrt {{n^2} - 3n - 5}  - \sqrt {9{n^2} + 3} }}{{2n - 1}} \\= \lim \frac{{( {\sqrt {{n^2} - 3n - 5}  - \sqrt {9{n^2} + 3} } ).( {\sqrt {{n^2} - 3n - 5}  + \sqrt {9{n^2} + 3} })}}{{( {\sqrt {{n^2} - 3n - 5}  + \sqrt {9{n^2} + 3} }).(2n - 1)}}\\ = \lim \frac{{({n^2} - 3n - 5) - (9{n^2} + 3)}}{{( {\sqrt {{n^2} - 3n - 5}  + \sqrt {9{n^2} + 3} } ).(2n - 1)}} \\= \lim \frac{{ - 8{n^2} - 3n - 8}}{{( {\sqrt {{n^2} - 3n - 5}  + \sqrt {9{n^2} + 3} } ).(2n - 1)}}\\ = \lim \frac{{ - 8 - \frac{3}{n} - \frac{8}{{{n^2}}}}}{{( {\sqrt {1 - \frac{3}{n} - \frac{5}{{{n^2}}}}  + \sqrt {9 + \frac{3}{{{n^2}}}} })( {2 - \frac{1}{n}} )}} = \frac{{ - 8}}{{4.2}} =  - 1.\end{array}$

Cách 2: Chia cả tử và mẫu cho $n.$

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