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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?

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  • $\begingroup$ Probably a typo for one of them, but the title says $2^{x^2}$, the problem says $2^{x^x}$, $\endgroup$ – WaveX Mar 1 '18 at 1:17
  • $\begingroup$ My mistake, sorry about that $\endgroup$ – Mathingmatics Mar 1 '18 at 1:18
  • $\begingroup$ Intuition: Since $x^2$ grow faster than $x$ then $a^{x^2}$ will overpower $b^x$ so the limit is $0$. That can probably by verified by taking logarithms. $\endgroup$ – fleablood Mar 1 '18 at 1:40
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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 $\displaystyle\lim_{x \to \infty} \dfrac {a^x}{2^{x^2}} =0$

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  • $\begingroup$ @Dosidis Ah thank you $\endgroup$ – Ovi Mar 1 '18 at 1:22
  • $\begingroup$ This answer is great, but it got me thinking - and I hope it isn't too off-topic for this thread - is this true for all values in place of $2000000$ that are greater than 1? If so, how would that be proven? I'll be accepting the answer in two minutes because of the time minimum. $\endgroup$ – Mathingmatics Mar 1 '18 at 1:27
  • $\begingroup$ @Mathingmatics Thank you for accepting my answer. It would work for any number instead of 2000000; I will post an edit to my answer in some minutes explaining it. $\endgroup$ – Ovi Mar 1 '18 at 1:37
  • $\begingroup$ For any $N$ there is going to be a $2^m > N$ so ... yes. $\endgroup$ – fleablood Mar 1 '18 at 1:41
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    $\begingroup$ @Mathingmatics I've added the edit. $\endgroup$ – Ovi Mar 1 '18 at 1:48
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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 $L=e^{\ln(L)} = e^{-\infty}=0$

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Hint: $$L=\lim_\limits{x\to\infty} \frac{1}{\left(\frac{2^x}{2000000}\right)^x}.$$

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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 $$ \lim_{x\to\infty}\frac{a^x}{b^{x^2}}=\lim_{x\to \infty}\exp(x\log a-x^2\log b)=\exp(\lim_{x\to \infty}(x\log a-x^2\log b))=0 $$

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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 $\ln(f(x)) \to -\infty$, $f(x) \to 0$.

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