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I want to show that the below limit is 0 for any polynomial $p(x)$ with degree $n$

$$\lim_{x \rightarrow \infty } \frac{p(x)}{2^{\sqrt x}} = 0$$

If I apply the l'Hopital's Rule, the numerator, eventually, will be zero. What about the denumerator?

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    $\begingroup$ Make the substitution $x=u^2$. $\endgroup$ – Szeto Oct 21 '18 at 1:41
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Suppose $p(x)$ is of $n$th degree and has coefficient of $x^n$ equals to $c$.

By the substitution $x=u^2$, $$\begin{align} \lim_{x\to\infty}\frac{p(x)}{2^{\sqrt x}} &=\lim_{u\to\infty}\frac{p(u^2)}{2^u}\\ &=\lim_{u\to\infty} \frac{p(u^2)}{cu^{2n}}\frac{cu^{2n}}{2^u} \\ &=\lim_{u\to\infty} \frac{p(u^2)}{cu^{2n}}\lim_{u\to\infty}\frac{cu^{2n}}{2^u} \\ &=1\cdot \lim_{u\to\infty}\frac{\frac{d^{2n}}{du^{2n}}cu^{2n}}{\frac{d^{2n}}{du^{2n}} 2^u} \\ &=\lim_{u\to\infty}\frac{c\cdot (2n)!}{(\ln 2)^{2n}2^u}\\ &=0 \end{align} $$

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We can assume the coefficient associated to $x^n$ is $1$. Hence, there exists $N>0$ such that $p(x)\geq0$, for all $x\geq N$. For $x\geq \max\{N,(n+1)^2\}$, we have $$ 0\leq \frac{p(x)}{x^{\sqrt x}}\leq \frac{p(x)}{x^{n+1}} $$ and $\lim_{x\to\infty}p(x)/x^{n+1}=0$. Therefore, $\lim_{x\to\infty}p(x)/x^{\sqrt{x}}=0$.

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Using the Binomial Theorem and no calculus. Let $x=y^2.$ Let deg$(p)=k.$ Then $p(y^2)$ is of degree $2k$ in y. Let $[y]$ denote the largest integer not exceeding $y.$ Note that $y\geq [y]>y-1.$

Now $$x\geq \sqrt {2k+1} \implies y\geq2k+1 \implies y\geq [y]\geq 2k+1 \implies$$ $$\implies 2^{\sqrt x}=2^y\geq 2^{[y]}=(1+1)^{[y]}=\sum_{k=0}^{[y]}\binom {[y]}{j}>$$ $$>\binom {[y]}{2k+1}=$$ $$=(2k+1)!^{-1}\prod_{i=0}^{2k}([y]-i)>$$ $$> (2k+1)!^{-1}\prod_{i=0}^{2k}(y-1-i).$$ Call the last expression above $q(y).$ Then $q(y)$ is a polynomial in $y $ of degree $2k+1$ while the degree of $p(y^2)$ is $2k.$

So $|p(y^2)/q(y)|\to 0$ as $y\to \infty.$

And for $x\geq \sqrt {2k+1}$ we have $|p(x)/2^{\sqrt x}|= |p(y^2)/2^y|<|p(y^2)/q(y)|. $

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  • $\begingroup$ This method works for a variety of similar formulas. $\endgroup$ – DanielWainfleet Oct 21 '18 at 8:02
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It suffices to show

$\lim_{x \rightarrow \infty}\dfrac{x^n}{2^{√x}} =0$ (why?).

1) Let $y=√x, y >0.$

Then $F(y)= \dfrac{y^{2n}}{2^y}.$

2) Let $e^a=2,$ $a >0.$

$F(y)= \dfrac{y^{2n}}{e^{ay}}.$

$e^{ay} \gt \dfrac{(ay)^{2n+1}}{(2n+1)!}$ (Series expansion).

$F(y) \lt \dfrac{(2n+1)! y^{2n}}{(ay)^{2n+1}}=$

$\dfrac{(2n+1)!}{a^{2n+1}y}=(\dfrac{(2n+1)!}{a^{2n+1}})\dfrac{1}{y}.$

Take the limit ${y \rightarrow \infty}.$

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  • $\begingroup$ on the last equation, $(2n+)!$ shuold be $(2n+1)!$ $\endgroup$ – kelalaka Oct 21 '18 at 9:11
  • $\begingroup$ kelalaka.Thanks:). $\endgroup$ – Peter Szilas Oct 21 '18 at 10:24
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Base on the nice hint;

$$\lim_{x \rightarrow \infty } \frac{p(x)}{2^\sqrt{x}} = \lim_{u^2 \rightarrow \infty } \frac{p(u^2)}{2^u} = \lim_{u^2 \rightarrow \infty } \frac{ \frac{d^{2n}} {du^{2n}} p(u^2)} {\frac{d^{2n}} {du^{2n}} 2^u} = \lim_{u^2 \rightarrow \infty } \frac{c}{2^{u} \log^{2n} 2}=0,$$ for some $c\in \mathbb{Z},$

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