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.
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Try to factorize
$x^n.$ -
… then take the limit.
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What is
$\lim_{x\to\infty}\frac{1}{x^k}$for any positive integer$k?$
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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.