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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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