Lời giải:
a) un=n23n2+7n−2
Ta có: lim = \mathop {\lim }\limits_{n \to + \infty } \frac{{{n^2}}}{{{n^2}\left( {3 + \frac{7}{n} - \frac{2}{{{n^2}}}} \right)}}
= \mathop {\lim }\limits_{n \to + \infty } \frac{1}{{3 + \frac{7}{n} - \frac{2}{{{n^2}}}}} = \frac{1}{3}.
b) {v_n} = \sum\limits_{k = 0}^n {\frac{{{3^k} + {5^k}}}{{{6^k}}}} = \frac{{{3^0} + {5^0}}}{{{6^0}}} + \frac{{{3^1} + {5^1}}}{{{6^1}}} + \frac{{{3^2} + {5^2}}}{{{6^2}}} + ... + \frac{{{3^n} + {5^n}}}{{{6^n}}}
= \left( {\frac{{{3^0}}}{{{6^0}}} + \frac{{{5^0}}}{{{6^0}}}} \right) + \left( {\frac{{{3^1}}}{{{6^1}}} + \frac{{{5^1}}}{{{6^1}}}} \right) + \left( {\frac{{{3^2}}}{{{6^2}}} + \frac{{{5^2}}}{{{6^2}}}} \right) + ... + \left( {\frac{{{3^n}}}{{{6^n}}} + \frac{{{5^n}}}{{{6^n}}}} \right)
= \left( {{{\left( {\frac{1}{2}} \right)}^0} + {{\left( {\frac{5}{6}} \right)}^0}} \right) + \left( {{{\left( {\frac{1}{2}} \right)}^1} + {{\left( {\frac{5}{6}} \right)}^1}} \right) + \left( {{{\left( {\frac{1}{2}} \right)}^2} + {{\left( {\frac{5}{6}} \right)}^2}} \right) + ... + \left( {{{\left( {\frac{1}{2}} \right)}^n} + {{\left( {\frac{5}{6}} \right)}^n}} \right)
= \left[ {{{\left( {\frac{1}{2}} \right)}^0} + {{\left( {\frac{1}{2}} \right)}^1} + {{\left( {\frac{1}{2}} \right)}^2} + ... + {{\left( {\frac{1}{2}} \right)}^n}} \right] + \left[ {{{\left( {\frac{5}{6}} \right)}^0} + {{\left( {\frac{5}{6}} \right)}^1} + {{\left( {\frac{5}{6}} \right)}^2} + ... + {{\left( {\frac{5}{6}} \right)}^n}} \right]
Vì {\left( {\frac{1}{2}} \right)^1} + {\left( {\frac{1}{2}} \right)^2} + ... + {\left( {\frac{1}{2}} \right)^n} là tổng n số hạng đầu của cấp số nhân với số hạng đầu là {\left( {\frac{1}{2}} \right)^1} = \frac{1}{2} và công bội là \frac{1}{2} nên
{\left( {\frac{1}{2}} \right)^0} + {\left( {\frac{1}{2}} \right)^1} + {\left( {\frac{1}{2}} \right)^2} + ... + {\left( {\frac{1}{2}} \right)^n} = {\left( {\frac{1}{2}} \right)^0} + \frac{{\frac{1}{2}\left( {1 - {{\left( {\frac{1}{2}} \right)}^n}} \right)}}{{1 - \frac{1}{2}}} = 1 + \left( {1 - {{\left( {\frac{1}{2}} \right)}^n}} \right) = 2 - {\left( {\frac{1}{2}} \right)^n}.
Tương tự, ta tính được:
{\left( {\frac{5}{6}} \right)^0} + {\left( {\frac{5}{6}} \right)^1} + {\left( {\frac{5}{6}} \right)^2} + ... + {\left( {\frac{5}{6}} \right)^n} = {\left( {\frac{5}{6}} \right)^0} + \frac{{\frac{5}{6}\left( {1 - {{\left( {\frac{5}{6}} \right)}^n}} \right)}}{{1 - \frac{5}{6}}} = 1 + 5\left( {1 - {{\left( {\frac{5}{6}} \right)}^n}} \right) = 6 - 5 \cdot {\left( {\frac{5}{6}} \right)^n}.
Do đó, {v_n} = \sum\limits_{k = 0}^n {\frac{{{3^k} + {5^k}}}{{{6^k}}}} = \left[ {2 - {{\left( {\frac{1}{2}} \right)}^n}} \right] + \left[ {6 - 5 \cdot {{\left( {\frac{5}{6}} \right)}^n}} \right] = 8 - {\left( {\frac{1}{2}} \right)^n} - 5 \cdot {\left( {\frac{5}{6}} \right)^n}.
Vậy \mathop {\lim }\limits_{n \to + \infty } {v_n} = \mathop {\lim }\limits_{n \to + \infty } \left( {\sum\limits_{k = 0}^n {\frac{{{3^k} + {5^k}}}{{{6^k}}}} } \right) = \mathop {\lim }\limits_{n \to + \infty } \left[ {8 - {{\left( {\frac{1}{2}} \right)}^n} - 5 \cdot {{\left( {\frac{5}{6}} \right)}^n}} \right] = 8.
c) {{\rm{w}}_n} = \frac{{\sin \,n}}{{4n}}
Ta có: \left| {{{\rm{w}}_n}} \right| = \left| {\frac{{\sin \,n}}{{4n}}} \right| \le \frac{1}{{4n}} < \frac{1}{n} và \mathop {\lim }\limits_{n \to + \infty } \frac{1}{n} = 0.
Do đó, \mathop {\lim }\limits_{n \to + \infty } {{\rm{w}}_n} = \mathop {\lim }\limits_{n \to + \infty } \frac{{\sin \,n}}{{4n}} = 0.