Tìm số thực a khác 0 sao cho \(\mathop {\lim }\limits_{n \to + \infty } \frac{{{n^2} - 2}}{{a{n^2} - 1}} = 2\)
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A.
\(a = - \frac{1}{2}\) .
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B.
\(a = - 2\).
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C.
\(a = 2\).
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D.
\(a = \frac{1}{2}\).
Sử dụng kiến thức giới hạn dãy số: Nếu \(\mathop {\lim }\limits_{n \to + \infty } {u_n} = a,\mathop {\lim }\limits_{n \to + \infty } {v_n} = b \ne 0\) thì \(\mathop {\lim }\limits_{n \to + \infty } \frac{{{u_n}}}{{{v_n}}} = \frac{a}{b}\).
Ta có: \(\mathop {\lim }\limits_{n \to + \infty } \frac{{{n^2} - 2}}{{a{n^2} - 1}} = \mathop {\lim }\limits_{n \to + \infty } \frac{{{n^2}\left( {\frac{{{n^2}}}{{{n^2}}} - \frac{2}{{{n^2}}}} \right)}}{{{n^2}\left( {\frac{{a{n^2}}}{{{n^2}}} - \frac{1}{{{n^2}}}} \right)}} = \mathop {\lim }\limits_{n \to + \infty } \frac{{1 - \frac{2}{{{n^2}}}}}{{a - \frac{1}{{{n^2}}}}} = \frac{1}{a}\)
Do đó, \(\frac{1}{a} = 2 \Rightarrow a = \frac{1}{2}\left( {tm} \right)\)
Đáp án : D