Computing $\lim_{x \to \infty} \frac{|x^n|}{e^x}$ How to prove that $e^x$ goes faster to infinity than any polynomial of $x$ without using the Taylor expansion of $e^x$ or L'hopital rule? in other words, the proof that:
$$\lim_{x \to \infty}  |x^n|e^{-x}=0$$
I tried to bound the expression from above by a function greater than $|x^n|$ for $x$ greater than some $\delta$ to apply squeeze theorem. I tried proving the limit directly, but both times I could find no excuse for the existence of such $\delta$ without using the known Taylor expansion.
 A: You can use the root test for series: $$ \sum_{k=1}^{\infty} \frac{k^n}{e^{k}} $$ and check that: $$\sqrt[k]{\frac{k^n}{e^k}} = \frac{\sqrt[k]{k}^n}{e} \stackrel{k\to\infty}{\longrightarrow} \frac{1}{e} < 1$$
so the series is convergent and (using necessary condition):
$$\lim_{k\to\infty} k^n e^{-k} = 0 $$
A: Set $x=nt$, so the limit you have to compute is
$$
\lim_{t\to\infty}\frac{n^nt^n}{(e^t)^n}=
\lim_{t\to\infty}n^n\biggl(\frac{t}{e^t}\biggr)^n
$$
and, as $n^n$ is constant, this is the same as proving that
$$
\lim_{t\to\infty}\frac{t}{e^t}=0.
$$
With the substitution $t=-\log u$, this becomes proving that
$$
\lim_{u\to0^+}-u\log u=0.
$$
By definition,
$$
\log u=\int_{1}^u\frac{1}{v}\,dv=-\int_{u}^1\frac{1}{v}\,dv
$$
(we can assume $0<u<1$, of course).
Consider the subdivision of the interval $[u,1]$ into $[u,\sqrt{u}]$ and $[\sqrt{u},1]$ and the lower Riemann sum corresponding to it:
$$
(\sqrt{u}-u)\frac{1}{\sqrt{u}}+(1-\sqrt{u})\frac{1}{1}=2(1-\sqrt{u}).
$$
The upper Riemann sum is
$$
(\sqrt{u}-u)\frac{1}{u}+(1-\sqrt{u})\frac{1}{\sqrt{u}}=
2\frac{1-\sqrt{u}}{\sqrt{u}}.
$$
Therefore
$$
2(1-\sqrt{u})\le -\log u\le2\frac{1-\sqrt{u}}{\sqrt{u}}
$$
so that
$$
2u(1-\sqrt{u})\le -u\log u\le2\sqrt{u}(1-\sqrt{u}).
$$
By squeezing,
$$
\lim_{u\to0}u\log u=0.
$$
A: $$x^n = e^{\ln{x^n}} = e^{n\ln x}.$$
So,
$$x^n e^{-x} = e^{n\ln x-x} \rightarrow 0$$
Since the exponent $\rightarrow -\infty$, when $x\rightarrow \infty$, which is easy to prove.
