Bài tập 17 trang 143 SGK Toán 11 NC

a) \(\lim \left( {3{n^3} - 7n + 11} \right) = \lim {n^3}\left( {3 - \frac{7}{{{n^2}}} + \frac{{11}}{{{n^3}}}} \right) =  + \infty \)

(vì \(\lim {n^3} =  + \infty ,\lim \left( {3 - \frac{7}{{{n^2}}} + \frac{{11}}{{{n^3}}}} \right) = 3 > 0\))

b) \(\lim \sqrt {2{n^4} - {n^2} + n + 2}  = \lim {n^2}.\sqrt {2 - \frac{1}{{{n^2}}} + \frac{1}{{{n^3}}} + \frac{2}{{{n^4}}}}  =  + \infty \)

(vì \(\lim {n^2} =  + \infty ,\lim \sqrt {2 - \frac{1}{{{n^2}}} + \frac{1}{{{n^3}}} + \frac{2}{{{n^4}}}}  = \sqrt 2  > 0\))

c) \(\lim \sqrt[3]{{1 + 2n - {n^3}}} = \lim n.\sqrt[3]{{\frac{1}{{{n^3}}} + \frac{2}{{{n^2}}} - 1}} =  - \infty \) 

(vì \(\lim n =  + \infty ,\lim \sqrt[3]{{\frac{1}{{{n^3}}} + \frac{2}{{{n^2}}} - 1}} =  - 1 < 0\))

d) Ta có \(\sqrt {{{2.3}^n} - n + 2}  = {\left( {\sqrt 3 } \right)^n}.\sqrt {2 - \frac{n}{{{3^n}}} + \frac{2}{{{3^n}}}} \) với mọi n.

Vì \(\lim \frac{n}{{{3^n}}} = 0\) (kết quả bài 4) và \(\lim \frac{2}{{{3^n}}} = 0\) nên \(\lim \sqrt {2 - \frac{n}{{{3^n}}} + \frac{2}{{{3^n}}}}  = \sqrt 2  > 0\)

Ngoài ra \(\lim {\left( {\sqrt 3 } \right)^n} =  + \infty \) 

Do đó \(\lim \sqrt {{{2.3}^n} - n + 2}  =  + \infty \).

-- Mod Toán 11 HỌC247