# Do polynomials have a limit ratio of 1 over fixed intervals?

If $$p$$ is an increasing polynomial function, and $$c>0$$ is a constant, do we always have: $$\underset{x \rightarrow \infty}{\lim} \dfrac{p(xc+c)}{p(xc)} = 1?$$

I think if I am not missing anything, as the coefficient and degree of the leading terms will be equal in the numerator and denominator, by L'Hôspital's rule the limit should be equal to $$1$$. I am wondering if there is an easier proof for this fact, or anything that I am missing.

• you should start with $t= xc$ and $P(t+c_1)/ P(t)$ Aug 26 '21 at 18:40
• Thank you @WillJagy ! I agree it is nicer to first substitute $t$ and then take the limit over $t$. Aug 26 '21 at 18:48
• Alt. hint: if $p$ has degree $n\ge 1$ then $p(t+c)-p(t) = r(t)$ is a polynomial of degree $\le n-1$. Then $p(t+c)/p(t) = 1+r(t)/p(t)$ where the second term has limit $0$.
– dxiv
Aug 27 '21 at 4:57

## 5 Answers

You don't need L'Hopital's Rule to prove this. If$$p(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0,$$with $$N\in\Bbb N$$ and $$a_n\ne0$$, then\begin{align}\lim_{x\to\infty}\frac{p(xc+c)}{p(xc)}&=\lim_{x\to\infty}\frac{p\bigl((x+1)c\bigr)}{p(xc)}\\&=\lim_{x\to\infty}\frac{a_n\bigl((x+1)c\bigr)^n+a_{n-1}\bigl((x+1)c\bigr)^{n-1}+\cdots+a_1\bigl((x+1)c\bigr)+a_0}{a_n(xc)^n+a_{n-1}(xc)^{n-1}+\cdots+a_1(xc)+a_0}\\&=\lim_{x\to\infty}\frac{a_nc^nx^n+\text{terms of smaller degree}}{a_nc^nx^n+\text{terms of smaller degree}}\end{align}and this limit is equal to $$1$$; just divide both the numerator and the denominator by $$x^n$$. Then it's clear that the limit is $$\frac{a_nc^n}{a_nc^n}=1$$.

• To be clearer (pedagogically), I'd suggest explicitly stating that the final fraction is referring to terms of smaller degree in $x$. And I'd just divide by $x^n$ to avoid having to note that neither $a_n$ nor $c$ can be $0$. Aug 26 '21 at 18:52
• @RobertShore I have edited my answer. Aug 26 '21 at 18:54
• @JoséCarlosSantos thanks for this answer! Is it correct that we did not assume $p$ is increasing? And can we use the small-oh notation for "terms of small degree", e.g., $o(x^n)$? Sep 26 '21 at 19:27
• Indeed , I made no assumption about whether or not $p(x)$ is increasing. And, yes, it is correct to use $o(x^n)$ instead of “terms of smaller degree”. Sep 26 '21 at 21:58
• @JoséCarlosSantos thank you! Because small-oh notation has some absolute value in the definition I was confused, but this is very clear. Thanks!! Sep 26 '21 at 23:45

It is true what you say. Recall that if $$P(x)$$ and $$Q(x)$$ are two polynomials of the same degree, then $$\displaystyle \lim_{x\to \pm \infty}\dfrac{P(x)}{Q(x)}=\dfrac{a_n}{b_n}$$, with $$a_n$$ and $$b_n$$ being the leading coefficients of $$P(x)$$ and $$Q(x)$$, respectively.

In the case we are considering, the two polynomials have the same degree and $$a_n=kc^n=b_n$$, where $$k$$ is the principal coefficient of $$p$$. Therefore,

$$\lim_{x\to \infty}\dfrac{p(xc+c)}{p(xc)}=\dfrac{kc^n}{kc^n}=1.$$

Yes, there's an easier proof. Divide the numerator and denominator of the fraction by $$x^n$$, where $$n = \deg p$$. Then your numerator and denominator both end up as $$a_nc^n$$ (where $$a_n$$ is the coefficient of the $$x^n$$ term of your polynomial) plus a bunch of terms that go to $$0$$ as $$x \to \infty$$. By the definition of degree, $$a_n \neq 0$$, and you're given that $$c \neq 0$$ (in fact, $$c \gt 0$$), so this fraction goes to $$1$$.

A polynomial can be written as: $$a(x-\lambda_1)(x-\lambda_2)\dots (x-\lambda_n)$$ where $$\lambda_1,\lambda_2\dots\lambda_n \in \mathbb{C}$$

The leading coeficient $$a$$ factors out of the ratio, which "distributes" over the sub-polynomials, so it is sufficient to prove $$\lim_{x\to \infty}\dfrac{p(xc+c)}{p(xc)}=1$$ for $$p(x) = (x-\lambda_{k})$$ . We get $$\frac{xc + c-\lambda_{k}}{xc-\lambda_{k}} = 1 + \frac{c}{xc-\lambda_{k}} \to 1$$ as $$x \to \infty$$.

It suffices to show that for any non zero polynomial with degree $$n$$

$$\lim_{x\to \infty}\frac{p(x)}{x^n}=a_n$$

then

$$\lim_{x\to \infty}\frac{p(xc+c)}{p(xc)}=\lim_{x\to \infty}\frac{p(xc+c)}{(xc+c)^n}\frac{(xc)^n}{p(xc)}\frac{(xc+c)^n}{x^n}\frac{x^n}{(xc)^n}=\frac{a_nc^n}{a_nc^n}=1$$