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Considering that asymptotically, $2^x$ grows faster than $x^5$ (in the beginning, $x^5$ grows faster than $2^x$, but there will be a point where $2^x$ outgrows $x^5$) then $\dfrac{x^5}{2^x} \rightarrow 0$ as $x \rightarrow \infty$. Therefore,
$$\lim _{x\to \infty}\dfrac{x^5}{2^x} = 0$$

But in order to solve the limit, I applied L'Hospital's Rule five times

\begin{align} \lim _{x\to \infty}\dfrac{x^5}{2^x} & =\lim _{x\to \infty}\dfrac{5x^4}{2^x\ln 2}\\ & = \lim _{x\to \infty}\frac{20x^3}{\ln^2(2)\cdot 2^x} \\ & = \lim _{x\to \infty}\frac{60x^2}{\ln^3(2)\cdot 2^x} \\ & = \lim _{x\to \infty}\frac{120x}{\ln^4(2)\cdot 2^x} \\ & = \lim _{x\to \infty}\frac{120}{\ln^5(2)\cdot 2^x} \\ & = \frac{120}{\ln^5(2)}\cdot\lim _{x\to \infty}\frac{1}{2^x} \\ & = 0 \end{align}

What would be a more elegant way solve it without using L'Hospital's Rule?

Edit

Even though, the Limit: $\lim_{n\to \infty} \frac{n^5}{3^n}$ is similar, I found the link provided by Axion004, How to prove that exponential grows faster than polynomial? more interesting. Also, the answer provided by user trancelocation was very interesting and is what I was expecting.

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    $\begingroup$ I think this answer is the easiest way to see this. Your question is an abstract duplicate of that question. $\endgroup$
    – Axion004
    Oct 1, 2021 at 16:56
  • $\begingroup$ There are many similar to your question. See here $\endgroup$
    – user947346
    Oct 1, 2021 at 18:28
  • $\begingroup$ @Axion004 Thank you, this link is really useful. $\endgroup$ Oct 1, 2021 at 18:38

7 Answers 7

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You can use the series expansion $e^t = \sum_{n=0}^{\infty}\frac{t^n}{n!}$ as follows:

For $x>0$ you have $$\frac{x^5}{2^x}= \frac{x^5}{e^{x\ln 2}}< \frac{x^5}{\frac{(x\ln 2)^6}{6!}}= \frac{6!}{\ln^6 2}\cdot \frac 1x$$

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Hint :

\begin{align*} \frac{x^5}{2^x} &= \exp \left(5 \ln(x)-x \ln(2)\right) \\ &= \exp \left[x\left(5 \frac{\ln(x)}{x}-\ln(2) \right)\right] \end{align*}

Now, you have the very classical limit (which can be proved with elementary method) $$\lim_{x \rightarrow +\infty} \frac{\ln(x)}{x} = 0$$

so $$\lim_{x \rightarrow +\infty} \left(5 \frac{\ln(x)}{x}-\ln(2) \right) = -\ln(2)$$

so $$\lim_{x \rightarrow +\infty} \left[x\left(5 \frac{\ln(x)}{x}-\ln(2) \right)\right] = -\infty$$

and you are done.

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    $\begingroup$ But then how do you show $5 \ln(x) - x \ln(2) \to - \infty$ without converting back to $x^5/2^x$? $\endgroup$
    – user56202
    Oct 1, 2021 at 15:03
  • $\begingroup$ @TheSilverDoe Thank you for the answer, but could you explain a bit further? I found intuitive that $\dfrac{x^5}{2^x} \rightarrow 0$ as $x \rightarrow \infty$, but I also would like to show, write down the process to get the solution. I am not sure how the hint would help. $\endgroup$ Oct 1, 2021 at 15:13
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    $\begingroup$ @MarkViola Don't you think it's true that if $f(x) \to \infty$, $g(x) \to L \in \mathbb{R} - \{0 \}$, then $f(x) \cdot g(x) \to \text{sgn}(L) \cdot \infty$? $\endgroup$
    – user56202
    Oct 1, 2021 at 16:40
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    $\begingroup$ @TheSilverDoe Thank you for the update. $\endgroup$
    – user56202
    Oct 1, 2021 at 16:41
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    $\begingroup$ @MarkViola In this answer, $f(x) = x$ and $\displaystyle g(x) = 5 \frac {\ln(x)}{x} - \ln(2)$. $f(x) \to +\infty$ and $g(x) \to - \ln(2) < 0$, so $f(x)g(x) \to - \infty$, so $e^{f(x)g(x)} \to 0$. Is there anything wrong with this? $\endgroup$
    – user56202
    Oct 2, 2021 at 21:49
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As an alternative way, by ratio test

$$\frac{\dfrac{(n+1)^5}{2^{n+1}}}{\dfrac{n^5}{2^n}}=\frac12\left(1+\frac1n\right)^5 \to \frac12 \implies \dfrac{n^5}{2^n} \to 0$$

and since $\forall x>0\quad \exists n$ such that $n\le x\le n+1$ we have

$$\dfrac{x^5}{2^x}\le \dfrac{(n+1)^5}{2^{n}}=2 \dfrac{(n+1)^5}{2^{n+1}} \to 0$$

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    $\begingroup$ (+1) Nicely done $\endgroup$
    – Mark Viola
    Oct 1, 2021 at 22:19
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    $\begingroup$ @MarkViola Thanks Mark! Nice to see you here around. Bye $\endgroup$
    – user
    Oct 2, 2021 at 7:47
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Let $x=5u$. Then

$$\lim_{x\to\infty}{x^5\over2^x}=5^5\left(\lim_{u\to\infty}{u\over2^u}\right)^5$$

so it suffices to compute $\lim_{u\to\infty}u/2^u$. Let's do this using an inequality starting with the binomial theorem:

$$2^n=(1+1)^n=1+{n\choose1}+{n\choose2}+\cdots+1\gt{n\choose2}={n(n-1)\over2}\ge{n^2\over4}$$

for integers $n\ge2$. It follows that

$$0\le{u\over2^u}\le{\lceil u\rceil\over2^{\lfloor u\rfloor}}\le{\lfloor u\rfloor+1\over\lfloor u\rfloor^2/4}=4\left({1\over\lfloor u\rfloor}+{1\over\lfloor u\rfloor^2}\right)\to0$$

so by the Squeeze Theorem, $\lim_{u\to\infty}u/2^u=0$.

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  • $\begingroup$ Hi, but isn't there a missing $5^5$ out of the limit sign? $\endgroup$
    – Sebastiano
    Oct 3, 2021 at 20:06
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    $\begingroup$ @Sebastiano, ah yes, you're right. I'll fix it. Thanks! $\endgroup$ Oct 3, 2021 at 20:08
  • $\begingroup$ Never mind I had voted with the assurance that you had written a very good answer. :-) $\endgroup$
    – Sebastiano
    Oct 3, 2021 at 20:09
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All we need is to know that $$ \lim_{t\to\infty}\frac{e^t}{t}=\infty \tag{1} $$

Let's prove that, for every $a>1$ and $b>0$ (not necessarily an integer), we have $$ \lim_{x\to\infty}\frac{x^b}{a^x}=0 \tag{2} $$ First of all, perform the substitution $x=by$, so $a^x=(a^y)^b$ and our limit becomes $$ \lim_{x\to\infty}b^b\Bigl(\frac{y}{a^y}\Bigr)^{b} \tag{3} $$ OK, if we can prove that the limit of the part in parentheses is $0$, we're done. It's quite similar to $(1)$, isn't it? Since $a^y=e^{y\log a}$, we can perform a further substitution $y\log a=z$ and we get $$ \lim_{y\to\infty}\frac{y}{a^y}=\lim_{z\to\infty}\frac{1}{\log a}\frac{z}{e^z} \tag{4} $$ which is indeed $0$ because of $(1)$. The assumption that $a>1$ has been used here, because in this case $\log a>0$.

Should we prove $(1)$? You find several proofs that don’t use l’Hôpital. Perhaps the simplest is to use the mean value theorem to prove that, for $t>0$, it holds that $$ e^t>1+t+\frac{t^2}{2} $$

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$x^5$ and $2^x$ are both increasing monotonically, this is important because it guarantees there's no "weird" behavior at any subset of the real line. At the same time, as you've noticed, $(2^x)'>(x^5)'$. This analysis guarantees the limit. Of course, to be rigorous about it, you'd have to prove both claims made here, so the L'hopital's rule solution might be the easiest method.

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    $\begingroup$ How does this help? $\endgroup$
    – Mark Viola
    Oct 1, 2021 at 16:35
  • $\begingroup$ @MarkViola It is a more elegant way to prove the limit of interest. $\endgroup$
    – Sigma
    Oct 1, 2021 at 17:30
  • $\begingroup$ It proved nothing of the kind. $\endgroup$
    – Mark Viola
    Oct 1, 2021 at 18:42
  • $\begingroup$ @MarkViola I agree. I just explained how it could be proved without l'hôpital. $\endgroup$
    – Sigma
    Oct 1, 2021 at 19:41
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    $\begingroup$ If, for large enough $x>0, x^5<2^x,$ why is it important that $x^5$ is monotonically increasing? At the very best, this should've been a comment. $\endgroup$ Oct 1, 2021 at 20:18
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In this answer it is shown without using L'Hopital's rule that for every $n>0$,

$$\lim_{x\to\infty}\frac{x^n}{e^x}=0. \tag{1}$$

We can use $(1)$ to show that for any $n>0$ and $a > 1$,

$$\lim_{x\to\infty}\frac{x^n}{a^x}=0. \tag{2}$$

To do this, write $a^x = e^{(\log a)x}$ where $\log a$ is positive since $a>1$. Then, if we set $y=(\log a)x$,

$$\frac{x^n}{a^x}=\frac{x^n}{e^{(\log a)x}}=\frac{1}{(\log a)^n}\frac{y^n}{e^y}.$$

When $x\to\infty$, we know that $y\to\infty$ because $\log a >0$. Therefore the behavior of $x^n/a^x$ follows from that of $y^n/e^y$ which is zero by $(1)$. Hence your limit is zero as it is a special case of $(2)$ where $n=5$ and $a=2$.

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