Tìm các giá trị của tham số a để \(\mathop {\lim }\limits_{n \to + \infty } \left( {\sqrt {4{n^2} - 5n + 8} + a - 2n} \right) = 1\).
Sử dụng quy tắc về giới hạn của 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: \(\mathop {\lim }\limits_{n \to + \infty } \left( {\sqrt {4{n^2} - 5n + 8} + a - 2n} \right) = \mathop {\lim }\limits_{n \to + \infty } \frac{{\left( {\sqrt {4{n^2} - 5n + 8} + a - 2n} \right)\left( {\sqrt {4{n^2} - 5n + 8} - \left( {a - 2n} \right)} \right)}}{{\sqrt {4{n^2} - 5n + 8} - \left( {a - 2n} \right)}}\)
\( = \mathop {\lim }\limits_{n \to + \infty } \frac{{\left( {4{n^2} - 5n + 8} \right) - {{\left( {a - 2n} \right)}^2}}}{{\sqrt {4{n^2} - 5n + 8} - \left( {a - 2n} \right)}} = \mathop {\lim }\limits_{n \to + \infty } \frac{{4an - 5n + 8 - {a^2}}}{{\sqrt {4{n^2} - 5n + 8} - \left( {a - 2n} \right)}} = \mathop {\lim }\limits_{n \to + \infty } \frac{{4a - 5 + \frac{8}{n} - \frac{{{a^2}}}{n}}}{{\sqrt {4 - \frac{5}{n} + \frac{8}{{{n^2}}}} - \frac{a}{n} + 2}} = \frac{{4a - 5}}{4}\)
Để \(\mathop {\lim }\limits_{n \to + \infty } \left( {\sqrt {4{n^2} - 5n + 8} + a - 2n} \right) = 1\) thì \(\frac{{4a - 5}}{4} = 1 \Leftrightarrow 4a = 9 \Leftrightarrow a = \frac{9}{4}\)