Tìm các giới hạn sau:
a) \(\lim \left( {1 + 3n - {n^2}} \right)\);
b) \(\lim \frac{{{n^3} + 3n}}{{2n - 1}}\);
c) \(\lim \left( {\sqrt {{n^2} - n} + n} \right)\);
d) \(\lim \left( {{3^{n + 1}} - {5^n}} \right)\).
Sử dụng kiến thức về giới hạn vô cực để tính: Giả sử \(\lim {u_n} = + \infty \) và \(\lim {v_n} = a\)
Nếu \(a > 0\) thì \(\lim {u_n}{v_n} = + \infty \).
Nếu \(a
a) \(\lim \left( {1 + 3n - {n^2}} \right) = \lim \left[ {{n^2}\left( {\frac{1}{{{n^2}}} + \frac{3}{n} - 1} \right)} \right]\)
Ta có: \(\lim {n^2} = + \infty ,\lim \left( {\frac{1}{{{n^2}}} + \frac{3}{n} - 1} \right) = - 1
Do đó, \(\lim \left( {1 + 3n - {n^2}} \right) = \lim {n^2}\left( {\frac{1}{{{n^2}}} + \frac{3}{n} - 1} \right) = - \infty \)
b) \(\lim \frac{{{n^3} + 3n}}{{2n - 1}} = \lim \left[ {{n^2}.\frac{{1 + \frac{3}{{{n^2}}}}}{{2 - \frac{1}{n}}}} \right]\)
Ta có: \(\lim {n^2} = + \infty ,\lim \left( {\frac{{1 + \frac{3}{{{n^2}}}}}{{2 - \frac{1}{n}}}} \right) = \frac{1}{2} > 0\)
Do đó, \(\lim \frac{{{n^3} + 3n}}{{2n - 1}} = \lim {n^2}\frac{{1 + \frac{3}{{{n^2}}}}}{{2 - \frac{1}{n}}} = + \infty \)
c) \(\lim \left( {\sqrt {{n^2} - n} + n} \right) = \lim \left[ {n\left( {\sqrt {1 - \frac{1}{n}} + 1} \right)} \right]\)
Ta có: \(\lim n = + \infty ,\lim \left( {\sqrt {1 - \frac{1}{n}} + 1} \right) = 2 > 0\)
Do đó, \(\lim \left( {\sqrt {{n^2} - n} + n} \right) = \lim \left[ {n\left( {\sqrt {1 - \frac{1}{n}} + 1} \right)} \right] = + \infty \)
d) \(\lim \left( {{3^{n + 1}} - {5^n}} \right) = \lim \left\{ {{5^n}\left[ {3.{{\left( {\frac{3}{5}} \right)}^n} - 1} \right]} \right\}\)
Ta có: \(\lim {5^n} = + \infty ,\lim \left[ {3.{{\left( {\frac{3}{5}} \right)}^n} - 1} \right] = 3.0 - 1 = - 1
Do đó, \(\lim \left( {{3^{n + 1}} - {5^n}} \right) = \lim \left\{ {{5^n}\left[ {3.{{\left( {\frac{3}{5}} \right)}^n} - 1} \right]} \right\} = - \infty \)