# 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?

• Probably a typo for one of them, but the title says $2^{x^2}$, the problem says $2^{x^x}$, – WaveX Mar 1 '18 at 1:17
• My mistake, sorry about that – Mathingmatics Mar 1 '18 at 1:18
• 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. – fleablood Mar 1 '18 at 1:40

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 $\displaystyle\lim_{x \to \infty} \dfrac {a^x}{2^{x^2}} =0$

• @Dosidis Ah thank you – Ovi Mar 1 '18 at 1:22
• 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. – Mathingmatics Mar 1 '18 at 1:27
• @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. – Ovi Mar 1 '18 at 1:37
• For any $N$ there is going to be a $2^m > N$ so ... yes. – fleablood Mar 1 '18 at 1:41
• @Mathingmatics I've added the edit. – Ovi Mar 1 '18 at 1:48

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 $L=e^{\ln(L)} = e^{-\infty}=0$

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

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

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