Đề bài
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) \(\mathop {\lim }\limits_{n \to + \infty } \left( {3n - \sqrt {4{n^2} + 1} } \right).\)
Phương pháp giải - Xem chi tiết
+ 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 chi tiết
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) \(\mathop {\lim }\limits_{n \to + \infty } \left( {3n - \sqrt {4{n^2} + 1} } \right) = \mathop {\lim }\limits_{n \to + \infty } n\left( {3 - \sqrt {4 + \frac{1}{{{n^2}}}} } \right) = + \infty \)
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