Answer
$\mathop {\lim }\limits_{n \to \infty } \frac{{3 - {4^n}}}{{2 + 7\cdot{4^n}}} = - \frac{1}{7}$
Work Step by Step
We have
$\mathop {\lim }\limits_{n \to \infty } \frac{{3 - {4^n}}}{{2 + 7\cdot{4^n}}} = \mathop {\lim }\limits_{n \to \infty } \frac{{3 - {4^n}}}{{2 + 7\cdot{4^n}}}\frac{{1/{4^n}}}{{1/{4^n}}} = \mathop {\lim }\limits_{n \to \infty } \frac{{\frac{3}{{{4^n}}} - 1}}{{\frac{2}{{{4^n}}} + 7}}$.
By (iii) and (iv) of Theorem 2 we get
$\mathop {\lim }\limits_{n \to \infty } \frac{{\frac{3}{{{4^n}}} - 1}}{{\frac{2}{{{4^n}}} + 7}} = \frac{{\mathop {\lim }\limits_{n \to \infty } \frac{3}{{{4^n}}} - \mathop {\lim }\limits_{n \to \infty } 1}}{{\mathop {\lim }\limits_{n \to \infty } \frac{2}{{{4^n}}} + \mathop {\lim }\limits_{n \to \infty } 7}} = \frac{{3\mathop {\lim }\limits_{n \to \infty } \frac{1}{{{4^n}}} - \mathop {\lim }\limits_{n \to \infty } 1}}{{2\mathop {\lim }\limits_{n \to \infty } \frac{1}{{{4^n}}} + \mathop {\lim }\limits_{n \to \infty } 7}} = - \frac{1}{7}$.
So, $\mathop {\lim }\limits_{n \to \infty } \frac{{3 - {4^n}}}{{2 + 7\cdot{4^n}}} = - \frac{1}{7}$.
