What would be the limit of $\frac{2000000^{x}}{2^{x^2}}$ as $x$ approaches infinity? I'm interested in the limit of the fraction:
$\frac{2000000^{x}}{2^{x^2}}$ as $x$ approaches infinity. Since the limit results in an indeterminate fraction of $\frac{\infty}{\infty}$, I was thinking of using l'hopitals rule but I don't think the derivatives would help much as the exponential would remain. How would this be computed?

We have $2000000<2^{21} $, so  $\dfrac{2000000^{x}}{2^{x^2}}<\dfrac {(2^{21})^x}{2^{x^2}}= \dfrac {1}{2^{x^2-21x}}$. Can you take it from here?
Generalization: There is nothing special about $2000000$; it could have been any other positive number, including a number less than $1$ $($ having  (negative number$)^x$ gets a little more tricky.
We want to show $\displaystyle \lim_{x \to \infty} \dfrac {a^x}{2^{x^2}}=0$. No matter what $a$ is, we can always find a number $n$ such that $a < 2^n$. Then  $ \displaystyle 0 \le \displaystyle \dfrac {a^x}{2^{x^2}} \le \dfrac {(2^n)^x}{2^{x^2}} = \dfrac {1}{2^{x^2-nx}}$, so $0 \le \lim_{x \to \infty} \dfrac {a^x}{2^{x^2}} \le  \lim_{x \to \infty} \dfrac {1}{2^{x^2-nx}}=0$, so $\displaystyle\lim_{x \to \infty} \dfrac {a^x}{2^{x^2}} =0$
