Show$\lim\limits_{x \to \infty} \frac{2^{2x}}{5^{x-1}} = 0$ Exactly as the title says.
$$\lim\limits_{x \to \infty} \frac{2^{2x}}{5^{x-1}} = 0$$
I am at a loss for how to show this one. At first I thought of using L'Hopital's rule on the numerator and the denominator, but it doesn't make the problem any simpler. 
Any help would be much appreciated.

Hint:
$$\frac{2^{2x}}{5^{x-1}}=5\frac{2^{2x}}{5^x}=5\frac{4^{x}}{5^x}=5\left(\frac{4}{5}\right)^x $$
