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Limit of a polynomial

  1. Nov 15, 2004 #1


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    Question: How to show that the limit as x goes to infinity of a given polynomial diverge? We have

    [tex] \lim_{x \rightarrow \infty} a_nx^n + a_{n-1}x^{n-1}+...+a_1x+a_0[/tex]

    but cannot say the this limit is the sum of the limit of each term separetly because none of these limit exist. We cannot do this either

    [tex] \lim_{x \rightarrow \infty} a_nx^n + a_{n-1}x^{n-1}+...+a_1x+a_0 = \lim_{x \rightarrow \infty} x^n(a_n + \frac{a_{n-1}}{x}+...+\frac{a_1}{x^{n-1}}+\frac{a_0}{x^n}) [/tex]

    and say the the limit is the product of the limit because one of the limit does not exist. So what then?

    Thank you.
  2. jcsd
  3. Nov 16, 2004 #2


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    Let's take the case [tex]a_{n}>0[/tex]
    What you CAN show, using your last equation, is that there exist [tex]N_{0}[/tex] so that [tex]\frac{a_{0}}{x^{n}}\geq{-\frac{a_{n}}{2n}}[/tex] whenever [tex]x\geq{N}_{0}[/tex]
    Similarly, it exists [tex]N_{1}[/tex] so that:
    [tex]\frac{a_{1}}{x^{n-1}}\geq{-\frac{a_{n}}{2n}}[/tex] whenever [tex]x\geq{N}_{1}[/tex]
    And so on.
    Setting N equal to the maximum of these [tex]N_{i}[/tex] values, yields the inequality:
    [tex]x^{n}(a_{n}+++\frac{a_{0}}{x^{n}})\geq\frac{x^{n}a_{n}}{2}, x\geq{N}[/tex]
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