What does ${\lim\limits_{x\to\infty} \frac{f(x)}{f(-x)}=-1}$ imply about a polynomial $f(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_0$ 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_{n-1}/x+a_{n-2}/x^2+\cdots+a_0/x^n)} {(-x)^n(a_n+a_{n-1}/(-x)+a_{n-2}/(-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.