a

$$\begin{array}{rl}\underset{n\to \infty }{\lim }\frac{(2^n-6{)}^2}{2\cdot 4^n+1}& \stackrel{(1)}{=}\underset{n\to \infty }{\lim }\frac{(2^n{)}^2-2\cdot 2^n\cdot 6+6^2}{2\cdot 4^n+1}\\ & =\underset{n\to \infty }{\lim }\frac{4^n-12\cdot 2^n+36}{2\cdot 4^n+1}\\ & =\underset{n\to \infty }{\lim }\frac{2^{2n}-12\cdot 2^n+36}{2\cdot 2^{2n}+1}\\ & =\underset{n\to \infty }{\lim }\frac{2^{2n}}{2^{2n}}\cdot \frac{1-\frac{12}{2^n}+\frac{36}{2^{2n}}}{2+\frac{1}{2^{2n}}}\\ & \stackrel{(2)}{=}\frac{{\lim }_{n\to \infty }1-\frac{12}{2^n}+\frac{36}{2^{2n}}}{{\lim }_{n\to \infty }2+\frac{2}{2^{2n2}}}\\ & =\frac{1}{2}\end{array}$$