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

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