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.

$\bf hint: $$$ \frac{2^{2x}}{5^{x-1}} = 5\left( \frac45 \right)^x$$
