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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?

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5 Answers 5

up vote 3 down vote accepted

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).

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No, that's good. Factoring x in the exponent gives $\lim\limits_{x \to \infty} x(\ln(2) - x\ln(3)) = \infty \times - \infty $. Then you get $e^{-\infty} = 0$. Is that right? –  bobobobo Feb 9 '11 at 2:46
    
That's certainly one way to think about the exponent, although I'm not sure how rigourous multiplying infinities is (and by that I mean I'm not sure if it is, not that it isn't, since it makes very much sense). I would be more comfortable factoring out an $x^2$ from the exponent instead and seeing what happens. –  Arpon Feb 9 '11 at 2:52
    
Yeah, that was the only thing I was unsure about. But I considered it as just sign mults. I can't confirm if that's wrong. –  bobobobo Feb 9 '11 at 2:55
    
The direct way to go here is by passing through a 'limit of a composition is the composition of the limit'-type of argument. –  Gerben Feb 9 '11 at 10:08

How about comparison to $\displaystyle{\frac{2^x}{3^x}=\left(\frac{2}{3}\right)^x}$?

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This is nice also. So would it go $\lim\limits_{x \to \infty}(\frac{2}{3})^x\frac{1}{3^x} = 0 \times 0$. I know that $(\frac{2}{3})^x$ should go to $0$ for big x because the base of the exponent is less than 1, but what's the hard and fast rule that lets us say that? –  bobobobo Feb 9 '11 at 2:53
    
@bobobobo: You have $0\lt$ your expression $\leq \frac{2^x}{3^x}\to 0$, showing that your expression goes to zero. There are a number of ways to see that $a^x$ goes to zero when $0\lt a\lt 1$. Since I see you like logs from another comment, notice that $\log(a^x)=x\log(a)\to -\infty$ because $\log(a)\lt 0$. Or notice that $y\gt x\Rightarrow a^y/a^x=a^{y-x}<1$, so $a^x$ is decreasing and positive, hence must have a limit, and since $\lim a^x = \lim a^{x+1}=a\lim a^x$, this limit must be $0$. –  Jonas Meyer Feb 9 '11 at 3:01
    
+1 This is the way to go imo. Just notice that $(2/3)^x\to 0$ and notice that $3^{x^2}$ goes to infinity faster than $3^x$. –  Myself Feb 9 '11 at 12:22

Hint: $2^x \lt 3^x$ for sufficiently large $x$.

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Yes.. so $2^x << 3^{x^2}$ for sufficiently large x as well.. is that a good enough argument? –  bobobobo Feb 9 '11 at 2:13
    
That is not what I was going for! Only apply that to the numerator and see the fraction you get... –  Aryabhata Feb 9 '11 at 2:17
    
Ok, $2^x < 3^x$, so write $\lim\limits_{x \to \infty}\frac{3^x}{3^{x^2}} > \lim\limits_{x \to \infty}\frac{2^x}{3^{x^2}}$, then $\lim\limits_{x \to \infty}\frac{1}{3^x} \to 0$. So it works because we established an upper bound that goes to 0? Why is that good enough? How do we nail down the $3^x$ numerator'd equation to the $2^x$ numerator'd equation? –  bobobobo Feb 9 '11 at 2:30
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@bobobobo: Note $\frac{3^x}{3^{x^2}} = \frac{1}{3^{x^2-x}}$ and $x^2-x \rightarrow \infty$ and $x \rightarrow \infty$ –  user17762 Feb 9 '11 at 2:37

How about rewriting $\frac{2^x}{3^{x^2}}$ as $\frac{2^x}{3^x3^{x^2-x}}=(\frac{2}{3})^x\cdot\frac{1}{3^{x^2-x}}$?

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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.

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