$\left ( \frac{4^{x+2}+ 3^{x}}{4^{x-2}} \right )$
=$\left ( \frac{4^{x}4^{2} + 3^{x}}{\frac{4^{x}}{4^{2}}} \right )$
=$16 \left ( \frac{4^{x}*16+ 3^{x}}{4^{x}} \right )$
Taking $4^{x}$ common from numerator and denominator and then cancelling it.
=$16 \left ( \frac{16+ (\frac{3}{4})^{x}}{1} \right )$
$\therefore \textrm{value of } _{x\rightarrow \infty }^{lim} \left ( \frac{4^{x+2}+ 3^{x}}{4^{x-2}} \right ) = _{x\rightarrow \infty }^{lim} 16 \left ( \frac{16+ (\frac{3}{4})^{x}}{1} \right ) = 16 \left ( \frac{16+0}{1} \right ) = 16*16 = 256$