Bài tập 5.2 trang 78 SBT Toán 11 Tập 1 Kết nối tri thức

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 giải chi tiết Bài 5.2

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

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

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

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

-- Mod Toán 11 HỌC247

Nếu bạn thấy hướng dẫn giải Bài tập 5.2 trang 78 SBT Toán 11 Tập 1 Kết nối tri thức - KNTT HAY thì click chia sẻ