Answer
(a) $\mathop {\lim }\limits_{n \to \infty } \left( {{a_n} + {b_n}} \right) = 11$
(b) $\mathop {\lim }\limits_{n \to \infty } {a_n}^3 = 64$
(c) $\mathop {\lim }\limits_{n \to \infty } \cos \left( {\pi {b_n}} \right) = - 1$
(d) $\mathop {\lim }\limits_{n \to \infty } \left( {{a_n}^2 - 2{a_n}{b_n}} \right) = - 40$
Work Step by Step
(a) By (i) of Theorem 2 (Limit Laws for Sequences):
$\mathop {\lim }\limits_{n \to \infty } \left( {{a_n} + {b_n}} \right) = \mathop {\lim }\limits_{n \to \infty } {a_n} + \mathop {\lim }\limits_{n \to \infty } {b_n} = 4 + 7 = 11$
(b) By (ii) of Theorem 2 (Limit Laws for Sequences):
$\mathop {\lim }\limits_{n \to \infty } {a_n}^3 = \left( {\mathop {\lim }\limits_{n \to \infty } {a_n}} \right)\left( {\mathop {\lim }\limits_{n \to \infty } {a_n}} \right)\left( {\mathop {\lim }\limits_{n \to \infty } {a_n}} \right) = {\left( {\mathop {\lim }\limits_{n \to \infty } {a_n}} \right)^3} = {4^3} = 64$
(c) We may define a function $f$ such that $f\left( {{b_n}} \right) = \cos \left( {\pi {b_n}} \right)$. So, by Theorem 4:
$\mathop {\lim }\limits_{n \to \infty } \cos \left( {\pi {b_n}} \right) = \cos \left( {\mathop {\lim }\limits_{n \to \infty } \pi {b_n}} \right) = \cos \left( {\pi \mathop {\lim }\limits_{n \to \infty } {b_n}} \right) = \cos \left( {7\pi } \right) = - 1$
(d) By (i) and (ii) of Theorem 2 (Limit Laws for Sequences):
$\mathop {\lim }\limits_{n \to \infty } \left( {{a_n}^2 - 2{a_n}{b_n}} \right) = \mathop {\lim }\limits_{n \to \infty } {a_n}^2 - 2\mathop {\lim }\limits_{n \to \infty } {a_n}{b_n} = {\left( {\mathop {\lim }\limits_{n \to \infty } {a_n}} \right)^2} - 2\left( {\mathop {\lim }\limits_{n \to \infty } {a_n}} \right)\left( {\mathop {\lim }\limits_{n \to \infty } {b_n}} \right)$
$\mathop {\lim }\limits_{n \to \infty } \left( {{a_n}^2 - 2{a_n}{b_n}} \right) = {4^2} - 2\cdot4\cdot7 = - 40$
