Bài 5.3 trang 109 SGK Toán 11 tập 1 - Kết nối tri thức

a) \(\lim {u_n} = \lim \frac{{{n^2} + 1}}{{2n - 1}} = \lim \frac{{1 + \frac{1}{{{n^2}}}}}{{\frac{2}{n} - \frac{1}{{{n^2}}}}}\).

Ta có: \(\lim \left( {1 + \frac{1}{{{n^2}}}} \right) = 1\), \(\lim \left( {\frac{2}{n} - \frac{1}{{{n^2}}}} \right) = 0\).

Suy ra \({u_n} =  + \infty \).

b) \(\lim {v_n} = \lim \left( {\sqrt {2{n^2} + 1}  - n} \right) = \lim \frac{{2{n^2} + 1 - {n^2}}}{{\sqrt {2{n^2} + 1}  + n}}\)

\( = \lim \frac{{{n^2} + 1}}{{{n^2}\left( {\sqrt {\frac{2}{{{n^2}}} + \frac{1}{{{n^4}}}}  + \frac{1}{n}} \right)}} = \lim \frac{{1 + \frac{1}{{{n^2}}}}}{{\sqrt {\frac{2}{{{n^2}}} + \frac{1}{{{n^4}}}}  + \frac{1}{n}}} =  + \infty \).