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Limits at infinity

  1. May 18, 2015 #1
    http://tutorial.math.lamar.edu/Classes/CalcI/LimitsAtInfinityI.aspx

    According to the author, if ##c## is a real number and ##r## is a positive rational number then:
    $$\lim_{x →\infty} \frac{c}{x^r} = 0$$
    If ##x^r## is defined for ##x < 0## then:
    $$\lim_{x →- \infty} \frac{c}{x^r} = 0$$
    I understand why ##r## can't be irrational in case two. ##x^r## would not be defined.
    However, I can't see why ##r## can't be irrational in case one.
     
  2. jcsd
  3. May 18, 2015 #2

    Svein

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    Well, it can. Rewrite: [itex]\frac{c}{x^{r}}=\frac{c}{e^{r\ln(x)}} [/itex]. Then [itex]\lim_{x\rightarrow \infty} \frac{c}{x^{r}}= \lim_{x\rightarrow \infty} \frac{c}{e^{r\ln(x)}}= \lim_{x\rightarrow \infty} e^{\ln (c)-r\ln(x)} [/itex].
     
  4. May 18, 2015 #3

    HallsofIvy

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    Saying "for r rational" does NOT say it is not also true for r irrational. It may or may not be in that case.
     
  5. May 18, 2015 #4
    When is it not true for irrational ##r##?
     
  6. May 18, 2015 #5

    HallsofIvy

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    You still do not understand. I am not saying statement is or is not true for any irrational r. I am simply pointing out that the initial statement "According to the author, if c c is a real number and r r is a positive rational number then:
    [itex]\lim_{x\to\infty} \frac{c}{x^r}= 0[/itex] does NOT say anything about what happens if r is irrational!
     
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