# How to solve $\lim _{x\to \infty}\dfrac{x^5}{2^x}$ without L'Hospital's Rule [duplicate]

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.

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

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

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

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.

• But then how do you show $5 \ln(x) - x \ln(2) \to - \infty$ without converting back to $x^5/2^x$? Oct 1, 2021 at 15:03
• @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. Oct 1, 2021 at 15:13
• @MarkViola Don't you think it's true that if $f(x) \to \infty$, $g(x) \to L \in \mathbb{R} - \{0 \}$, then $f(x) \cdot g(x) \to \text{sgn}(L) \cdot \infty$? Oct 1, 2021 at 16:40
• @TheSilverDoe Thank you for the update. Oct 1, 2021 at 16:41
• @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? Oct 2, 2021 at 21:49

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

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

• (+1) Nicely done Oct 1, 2021 at 22:19
• @MarkViola Thanks Mark! Nice to see you here around. Bye
– user
Oct 2, 2021 at 7:47

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, $$\lim_{u\to\infty}u/2^u=0$$.

• Hi, but isn't there a missing $5^5$ out of the limit sign? Oct 3, 2021 at 20:06
• @Sebastiano, ah yes, you're right. I'll fix it. Thanks! Oct 3, 2021 at 20:08
• Never mind I had voted with the assurance that you had written a very good answer. :-) Oct 3, 2021 at 20:09

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 $$e^t>1+t+\frac{t^2}{2}$$

$$x^5$$ 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.

• How does this help? Oct 1, 2021 at 16:35
• @MarkViola It is a more elegant way to prove the limit of interest. Oct 1, 2021 at 17:30
• It proved nothing of the kind. Oct 1, 2021 at 18:42
• @MarkViola I agree. I just explained how it could be proved without l'hôpital. Oct 1, 2021 at 19:41
• 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. Oct 1, 2021 at 20:18

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 $$a=2$$.