${\lim\limits_{x\to\infty} \frac{f(x)}{f(x)}=1}$
imply about a polynomial $f(x)=a_nx^n+a_{n1}x^{n1}+\cdots+a_0$
with real coefficients? Prove your answer.

Try to factorize
$x^n.$

… then take the limit.

What is
$\lim_{x\to\infty}\frac{1}{x^k}$
for any positive integer$k?$

Factorizing
$x^n$
and knowing$\lim_{x\to\infty}\frac{1}{x^k}=0$
for any positive integer$k$
, we have:$$\begin{aligned} \lim\limits_{x\to\infty} \frac{f(x)}{f(x)} &= \lim\limits_{x\to\infty} \frac{x^n(a_n+a_{n1}/x+a_{n2}/x^2+\cdots+a_0/x^n)} {(x)^n(a_n+a_{n1}/(x)+a_{n2}/(x)^2+\cdots+a_0/(x)^n)} \\ &= \lim_{x\to\infty} \frac{x^n(a_n+0+0+\cdots+0)}{(x)^n(a_n+0+0+\cdots+0)} \\ &= (1)^n. \end{aligned} $$
If this limit is$1,$
then$n$
must be odd.