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Limits to infinity with natural logs

  1. Mar 11, 2012 #1
    1. The problem statement, all variables and given/known data

    I'm having trouble calculating these limits.


    2. Relevant equations

    none

    3. The attempt at a solution

    1. lim.......((lnx)^5)/(x)=
    x->infinity

    (((lnx)^5)/x)/(x/x)=

    (((lnx)^5)/x)/1=

    How do I calculate from here?

    2.lim .......(14lnx^2)/(6lnx^3)=
    x->infinity

    ((14lnx^2)/(x^3))/((6lnx^3)/(x^3))=

    ((14lnx)/(x))/(6lnx)=

    0/6lnx= 0

    Is this right or am I forgetting something?
     
  2. jcsd
  3. Mar 11, 2012 #2
    use Lhospital
     
  4. Mar 11, 2012 #3

    SammyS

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    You're going in circles... back to what you started with.

    As Punky said, use L'Hôpital's rule.
    In #2, what you have done is incorrect.

    Use properties of logarithms for #2. It works out quite simply.

    [itex]\log_{\,b}(x^a)=(a)\log_{\,b}(x)\,.[/itex]
     
  5. Mar 11, 2012 #4
    I'm not allowed to use L'Hopital, and I will change #2
     
  6. Mar 11, 2012 #5
    So would the second problem look something like this:

    28lnx/18lnx?
     
  7. Mar 11, 2012 #6

    SammyS

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    If you mean the limit of 28lnx/(18lnx), then yes, that's right.

    I'll think about #1 some more. Maybe someone else will help.
     
  8. Mar 11, 2012 #7
    ok so it will be ((28lnx)/x)/((18lnx)/x) = 0/0 = DNE?
     
  9. Mar 11, 2012 #8

    SammyS

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    A limit of the form 0/0 is indeterminate. It might exist or might not exist.

    Why are you bothering to multiply the numerator & denominator by 1/x ? (That accomplishes nothing!)

    What is [itex]\displaystyle \frac{\ln(x)}{\ln(x)}\,[/itex] for any x in the domain of the natural log ?
     
  10. Mar 11, 2012 #9
  11. Mar 12, 2012 #10

    SammyS

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    For #1.

    Do you know [itex] \displaystyle \lim_{x\to\infty} \frac{\ln(x)}{x}\ ?[/itex]

    If so then rewrite [itex] \displaystyle \frac{(\ln(x))^5}{x}\ \text{ as }\ \frac{(\ln(x))^5}{x^5}\ x^4\ .[/itex]
     
  12. Mar 12, 2012 #11

    SammyS

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    Yes !

    What does that leave you with?
     
  13. Mar 12, 2012 #12
    So basically the limit is then 28/18?
     
  14. Mar 12, 2012 #13

    SammyS

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    Yes.
     
  15. Mar 12, 2012 #14
    I think I know how to do the first one by applying the chain rule:

    limit to infinity

    ((lnx)^5)/(x)=

    (5(lnx)^4)*(1/x)/(x)=
    5(lnx)^4*0/(x)= 0

    Is that valid?
     
  16. Mar 12, 2012 #15
    Ignore my last post I got confused, this isn't a derivative
     
  17. Mar 12, 2012 #16

    SammyS

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    Look at post #10 .
     
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