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I got a problem. I don't know how to solve this kind of problem:

Find an integer $n_0$ that for each $n>n_0$ this inequation would be always true:

$$\dfrac{n^{4}-n^{2}+1}{n^{3}-n}>10000 $$

I don't know how to solve this kind of problems, and I hope you could help me understanding how to proceed step by step. Thank you in advance.

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First of all, do you have any idea why such a number is guaranteed to exist? If so, then use that idea. If not, then look at the most significant terms in those two polynomials, and think about what happens when $n$ gets large. –  TonyK Jan 26 '11 at 14:02
    
Finding any integer $n_0$ where the inequality is true, that is easy. Finding the smallest integer requires only slightly more work. –  Joshua Shane Liberman Jan 26 '11 at 15:04

2 Answers 2

up vote 3 down vote accepted

Hints:

$\dfrac{n^{4}-n^{2}+1}{n^{3}-n}>10000\Leftrightarrow n+\dfrac{1}{n^{3}-n}>10000 $

$0<\dfrac{1}{n^{3}-n}<1\Leftrightarrow n>1$

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Thank you so much. i would like to know if this method is right as well: (n^3-n)/(n^4-n^2+1) < (n^3-n)/n^4 < n^3/n^4 < 1/10000 Solving this n > 10000 –  frx08 Jan 26 '11 at 14:47
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The majoration $\frac{n^{3}-n}{n^{4}-n^{2}+1}<\frac{n^{3}-n}{n^{4}}$ is not correct ($n^{4}-n^{2}+1<n^{4}$) –  Américo Tavares Jan 26 '11 at 14:56
    
Yes right! what about (n^3 - n )/(n ^4 -2n^2) < (n^3)/((n^2)(n^2-2)) and the result would be (n^2-2)/n > 10000 –  frx08 Jan 26 '11 at 15:16
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It is a possible valid approach. –  Américo Tavares Jan 26 '11 at 15:34

Can you divide out the fraction to a polynomial and remainder? That will help.

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