# How to show that $\lim_{x \rightarrow \infty } \frac{p(x)}{2^{\sqrt x}} = 0$

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?

• Make the substitution $x=u^2$. – Szeto Oct 21 '18 at 1:41

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}

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, $$\lim_{x\to\infty}p(x)/x^{\sqrt{x}}=0$$.

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 $$|p(x)/2^{\sqrt x}|= |p(y^2)/2^y|<|p(y^2)/q(y)|.$$

• This method works for a variety of similar formulas. – DanielWainfleet Oct 21 '18 at 8:02

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 $${y \rightarrow \infty}.$$

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

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 $$c\in \mathbb{Z},$$