Bài 5.2 trang 78 SBT toán 11 – Kết nối tri thức: Tính các giới hạn sau: a) mathop lim limits_n -> + ∇ √n^2 + 2n – n – 2 ;

Đề bài/câu hỏi:

Tính các giới hạn sau:

a) \(\mathop {\lim }\limits_{n \to + \infty } \left( {\sqrt {{n^2} + 2n} – n – 2} \right);\)

b) \(\mathop {\lim }\limits_{n \to + \infty } \left( {2 + {n^2} – \sqrt {{n^4} + 1} } \right);\)

c) \(\mathop {\lim }\limits_{n \to + \infty } \left( {\sqrt {{n^2} – n + 2} + n} \right);\)

d) \(\mathop {\lim }\limits_{n \to + \infty } \left( {3n – \sqrt {4{n^2} + 1} } \right).\)

Hướng dẫn:

+ Nếu \(\mathop {\lim }\limits_{n \to + \infty } {u_n} = a\) và \(\mathop {\lim }\limits_{n \to + \infty } {v_n} = + \infty \) (hoặc \(\mathop {\lim }\limits_{n \to + \infty } {v_n} = – \infty \)) thì \(\mathop {\lim }\limits_{n \to + \infty } \frac{{{u_n}}}{{{v_n}}} = 0\)

+ Nếu \(\mathop {\lim }\limits_{n \to + \infty } {u_n} = a > 0\) và \(\mathop {\lim }\limits_{n \to + \infty } {v_n} = 0\) và \({v_n} > 0\) với mọi n thì \(\mathop {\lim }\limits_{n \to + \infty } \frac{{{u_n}}}{{{v_n}}} = + \infty \)

+ Nếu \(\mathop {\lim }\limits_{n \to + \infty } {u_n} = + \infty \) và \(\mathop {\lim }\limits_{n \to + \infty } {v_n} = a > 0\) thì \(\mathop {\lim }\limits_{n \to + \infty } {u_n}{v_n} = + \infty \)

Lời giải:

a) \(\mathop {\lim }\limits_{n \to + \infty } \left( {\sqrt {{n^2} + 2n} – n – 2} \right) = \mathop {\lim }\limits_{n \to + \infty } \frac{{ – 2n – 4}}{{\sqrt {{n^2} + 2n} + n + 2}} = \mathop {\lim }\limits_{n \to + \infty } \frac{{ – 2 – \frac{4}{n}}}{{\sqrt {1 + \frac{2}{n}} + 1 + \frac{2}{n}}} = – 1\)

b) \(\mathop {\lim }\limits_{n \to + \infty } \left( {2 + {n^2} – \sqrt {{n^4} + 1} } \right) = \mathop {\lim }\limits_{n \to + \infty } \frac{{4{n^2} + 3}}{{2 + {n^2} + \sqrt {{n^4} + 1} }} = \mathop {\lim }\limits_{n \to + \infty } \frac{{4 + \frac{3}{{{n^2}}}}}{{\frac{2}{{{n^2}}} + 1 + \sqrt {1 + \frac{1}{{{n^4}}}} }} = 2\)

c) \(\mathop {\lim }\limits_{n \to + \infty } \left( {\sqrt {{n^2} – n + 2} + n} \right) = \mathop {\lim }\limits_{n \to + \infty } n\left( {\sqrt {1 – \frac{1}{n} + \frac{2}{{{n^2}}}} + 1} \right) = + \infty \)

d) \(\mathop {\lim }\limits_{n \to + \infty } \left( {3n – \sqrt {4{n^2} + 1} } \right) = \mathop {\lim }\limits_{n \to + \infty } n\left( {3 – \sqrt {4 + \frac{1}{{{n^2}}}} } \right) = + \infty \)