TITLE: How to solve $\lim _{x\to \infty}\dfrac{x^5}{2^x} $ without L'Hospital's Rule
QUESTION [7 upvotes]: 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.

REPLY [3 votes]: 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, $\lim_{u\to\infty}u/2^u=0$.