$$g(z) = \frac{4z^6-7z^3}{(Z^2-4)^3} = \frac{4z^6-7z^3}{z^6-12z^4+48z^2-64}$$
we can divide $z^6$ on both numerator and denominator.
Then we can get
$$g(z) = \frac{4-7z^{-1}}{z-12z^{-2}+48z^{-4}-64z^{-6}}$$
Then as $z\to\infty $
$$\lim_{z\to\infty} f(z) = \frac{4 - 0}{1 - 0 + 0 - 0} = 4$$
