$\lim\limits_{x \to \infty}\frac{2^x}{3^{x^2}}$ Find
$$\lim\limits_{x \to \infty}\frac{2^x}{3^{x^2}}$$
I can only reason with this intuitively. since $3^{x^2}$ grows much faster than $2^x$ the limit as $x \to \infty$ of $f$ must be 0.
Is there a more rigorous way to show this?
 A: How about comparison to $\displaystyle{\frac{2^x}{3^x}=\left(\frac{2}{3}\right)^x}$?
A: Hint: $2^x \lt 3^x$ for sufficiently large $x$.
A: How about rewriting $\frac{2^x}{3^{x^2}}$ as $\frac{2^x}{3^x3^{x^2-x}}=(\frac{2}{3})^x\cdot\frac{1}{3^{x^2-x}}$?
A: Rewriting the fraction as $$\frac{e^{x \ln 2}}{e^{x^2 \ln 3}} = e^{x \ln 2 - x^2 \ln 3}$$ may also help you (although perhaps no more than the other answers given already).
A: After so many equivalent examples one more should not be missed because of its form. Here we get the numerator constant and keep the notation in powers of 2 and 3. Denote $ ß=\frac{\ln(2)}{\ln(3)} $ Then 
$$ \frac{2^x}{3^{x^2}}=3^{ßx-x^2}=3^{(ß/2)²-(x-\fracß2)^2} = \frac{2^{ß/4}}{3^{(x-ß/2)^2}} $$ where the numerator is constant. We see also, that the denominator is minimal if $x=ß/2$ and the whole expression is maximal by the exponent $ßx-x^2$ of the second term, its first derivative $ß-2x$ (which is zero at $x=ß/2$) and its second derivative $-2$ which is negative and shows, that the whole expression has a maximum there.
