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?
 A: 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 $\displaystyle\lim_{x \to \infty} \dfrac {a^x}{2^{x^2}} =0$
A: Suppose that the limit exists..... then $L = \lim_{x\to \infty}\frac{2000000^x}{2^{x^2}}$ so $\ln(L) = \lim_{x\to \infty} x\ln(2000000) - (x^2)\ln(2)=-\infty$ so $L=e^{\ln(L)} = e^{-\infty}=0$
A: Hint:
$$L=\lim_\limits{x\to\infty} \frac{1}{\left(\frac{2^x}{2000000}\right)^x}.$$
A: This is equivalent to asking about the limit 
$$
\lim_{x\to \infty}cx-dx^2=-\infty
$$
Where $c$ and $d$ are positive constants. Indeed, for $a,b>0$, we have
$$
\frac{a^x}{b^{x^2}}=\frac{\exp(x\log a)}{\exp(x^2\log b)}=\exp(x\log a-x^2\log b)
$$
now use continuity of the exponential and note that as long as $a,b>1$ we may set $c=\log a$ and $d=\log b$ to find 
$$
\lim_{x\to\infty}\frac{a^x}{b^{x^2}}=\lim_{x\to \infty}\exp(x\log a-x^2\log b)=\exp(\lim_{x\to \infty}(x\log a-x^2\log b))=0
$$
A: More generally
(since this problem
is absurdly specific),
for
$a>1, b>1, c>1$
we have
$\lim_{x \to \infty} \dfrac{a^x}{b^{x^c}}
=0
$.
Let
$f(x)=\dfrac{a^x}{b^{x^c}}
$.
Then
$\ln(f(x))
= x\ln a-x^c\ln(b)
= x^c(x^{1-c}\ln a-\ln(b))
$.
Since
$c > 1$,
$x^{1-c} \to 0$
so
$x^{1-c}\ln a-\ln(b)
\lt 0$
for all large enough $x$.
Since
$x^c \to \infty$,
$x^c(x^{1-c}\ln a-\ln(b))
\to -\infty$
Since
$\ln(f(x))
\to -\infty$,
$f(x) \to 0$.
