I am asked to prove that if $p(x)\in \mathbb R[x] $ with the condition that $\deg (p(x))$ is odd then $\lim_{x\to \infty} p(x) = -\lim_{x\to-\infty} p(x)$

My approach: I want to show that either $$\lim\limits_{x \to \infty} p(x)=\infty $$ or$$\lim\limits_{x \to \infty} p(x)=-\infty$$ then $$\lim\limits_{x \to -\infty} p(x)=-\infty $$ or $$\lim\limits_{x \to -\infty} p(x)=\infty $$ respectively. Is this correct, is this idea worth considering?

Thank you for help beforehand.

  • $\begingroup$ I'm putting a minus sign in the statement to make it true. :-) $\endgroup$ – Sammy Black Mar 19 '13 at 22:27
  • 1
    $\begingroup$ For proof, you may want to split into two cases, (i) lead coefficient positive; (ii) negative. (Once you have done the first, the second is trivial.) $\endgroup$ – André Nicolas Mar 19 '13 at 22:30

There is nothing wrong with your approach.

You can also consider

Hint: Prove $$\lim\limits_{x\to\infty}\frac{p(-x)}{p(x)}=-1$$


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