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.
 A: 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.
A: 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
$$\dfrac{x^5}{2^x}\le \dfrac{(n+1)^5}{2^{n}}=2 \dfrac{(n+1)^5}{2^{n+1}} \to 0$$
A: 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, $\lim_{u\to\infty}u/2^u=0$.
A: 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
$$
e^t>1+t+\frac{t^2}{2}
$$
A: You can use the series expansion $e^t = \sum_{n=0}^{\infty}\frac{t^n}{n!}$ as follows:
For $x>0$ you have
$$\frac{x^5}{2^x}= \frac{x^5}{e^{x\ln 2}}< \frac{x^5}{\frac{(x\ln 2)^6}{6!}}= \frac{6!}{\ln^6 2}\cdot \frac 1x$$
A: $x^5$ 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.
A: 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 $a=2$.
