Show$\lim\limits_{x \to \infty} \frac{2^{2x}}{5^{x-1}} = 0$

Question

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.

Answer 1

Hint:

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

Answer 2

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