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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.

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    $\begingroup$ you should start with $t= xc$ and $P(t+c_1)/ P(t) $ $\endgroup$
    – Will Jagy
    Aug 26 '21 at 18:40
  • $\begingroup$ Thank you @WillJagy ! I agree it is nicer to first substitute $t$ and then take the limit over $t$. $\endgroup$ Aug 26 '21 at 18:48
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    $\begingroup$ 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 $0$. $\endgroup$
    – dxiv
    Aug 27 '21 at 4:57
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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 $\frac{a_nc^n}{a_nc^n}=1$.

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    $\begingroup$ 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 $0$. $\endgroup$ Aug 26 '21 at 18:52
  • $\begingroup$ @RobertShore I have edited my answer. $\endgroup$ Aug 26 '21 at 18:54
  • $\begingroup$ @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., $o(x^n)$? $\endgroup$ Sep 26 '21 at 19:27
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    $\begingroup$ 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”. $\endgroup$ Sep 26 '21 at 21:58
  • $\begingroup$ @JoséCarlosSantos thank you! Because small-oh notation has some absolute value in the definition I was confused, but this is very clear. Thanks!! $\endgroup$ Sep 26 '21 at 23:45
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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,

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

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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 $1$.

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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 $x \to \infty$.

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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

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

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