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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.

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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): $$\lim_{k\to\infty} k^n e^{-k} = 0 $$

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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, $$ \lim_{u\to0}u\log u=0. $$

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$$x^n = e^{\ln{x^n}} = e^{n\ln x}.$$ 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.

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