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Limit Problem

  1. Oct 13, 2012 #1
    lim__[[t^3 ln(t)] / 3] + (t^3)/9
    t->0+

    ^ in case that's difficult to read here's this.

    Now what I don't understand is how this comes out to be 0. I would think that ln(t) as t approaches 0 would be -∞ (since ln(t) keeps getting bigger and bigger in the - direction as t approaches o). You would multiply that by 0 (since t^3 as t approaches 0 gets closer and closer to 0). 0 * -∞ should be indeterminate. 0 * -∞ over 3 (i.e., the term: [t^3 ln(t)] / 3]) should also be indeterminate. I understand how (t^3)/9 is 0. Is this problem meant to be solved with L'Hospital's Rule?

    P.S. I've used L'Hospital's rule on it already and it does come out to zero. I just want to check with the people here to see if that is whatI'm supposed to do.
     
  2. jcsd
  3. Oct 13, 2012 #2

    Zondrina

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    Yes, Hospital's rule is appropriate for this question. Something else you could've done was use the squeeze theorem.
     
  4. Oct 13, 2012 #3
    Alright thanks. I appreciate it.
     
  5. Oct 13, 2012 #4

    Ray Vickson

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    For any fixed power p > 0 we have
    [tex] \lim_{x \rightarrow 0+} x^p \ln(x) = 0,[/tex] so things like
    [tex] x^{1/10} \ln(x), \; \sqrt{x} \ln(x),\; x \ln(x), \ldots [/tex]
    all → 0 as x → 0+. It might help to write
    [tex] x^p \ln(x) = \ln \left(x^{x^p}\right) \text{ for } x > 0.[/tex]

    RGV
     
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