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Calculating limits to infinity

  1. Mar 10, 2012 #1
    1. The problem statement, all variables and given/known data

    Can someone tell if if these look right?


    2. Relevant equations
    none


    3. The attempt at a solution

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

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

    (5lnx/x)/1 = 0/1 = 0

    2. lim sinx/x^2=
    x->infinity

    (sinx/x^2)/(x^2/x^2)=

    (sinx/x*x)/1=

    (sinx/x)*(1/x)/1 = (sinx/x * 0)/1 = 0
     
  2. jcsd
  3. Mar 10, 2012 #2

    Dick

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    (lnx)^5 isn't equal 5*ln(x). That doesn't look right. What technique are you trying to use here? Or is there just a notational problem?
     
  4. Mar 10, 2012 #3
    I thought you could do that with natural logs?
     
  5. Mar 10, 2012 #4

    Dick

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    ln(x^5)=5ln(x). ln(x)^5 isn't equal to 5ln(x). They are two different things. If you meant the first thing you should use parentheses differently.
     
    Last edited: Mar 10, 2012
  6. Mar 10, 2012 #5
    Oh, so how would i calculate it then?
     
  7. Mar 11, 2012 #6
    Cacophony,

    apparently you meant:


    [tex]\displaystyle\lim_{x\to \infty} \ \bigg[ln(x)\bigg]^{\dfrac{5}{x}}[/tex]
     
  8. Mar 11, 2012 #7
    No, i meant ((lnx)^5)/(x)
     
  9. Mar 11, 2012 #8

    Mentallic

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    Yes, that's correct, but your line of (sinx/x^2)/(x^2/x^2) was unnecessary. You should immediately split up [tex]\frac{\sin(x)}{x^2}=\frac{\sin(x)}{x} \cdot \frac{1}{x}[/tex] as so.


    L'Hospital's rule? :smile:
     
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