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Practice Paper 2 Question 6

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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Hints

  • Hint 1
    Try to factorize \(x^n.\)
  • Hint 2
    ... then take the limit.
  • Hint 3
    What is \(\lim_{x\to\infty}\frac{1}{x^k}\) for any positive integer \(k?\)

Solution

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

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