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

Question

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.

Answer 1

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

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

Answer 2

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.

Answer 3

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

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

Answer 4

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

Answer 5

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

Answer 6

$x^5$ 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.

Answer 7

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 $a=2$.