Bài 5.2 trang 78 SBT Toán 11 - Kết nối tri thức: Tính các giới hạn sau: \(\mathop {\lim }\limits_{n \to + \infty } \left( {\sqrt {{n^2} + 2n} - n...

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

+ 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 $\mathop {\lim }\limits_{n \to + \infty } {u_n}{v_n} = + \infty$

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