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Limit. Does it exist or not?

  1. May 30, 2013 #1
    1. lim as x->o+ ( (sqrt(x+sinx))(lnx))



    2. Relevant equations: [PLAIN]http://www.math.oregonstate.edu/home/programs/undergrad/CalculusQuestStudyGuides/SandS/lHopital/Laws/multiplication_law.gif[/b] [Broken]



    3. The attempt at a solution:

    lim as x->o+ ( (sqrt(x+sinx))(lnx))
    = lim as x->o+ (sqrt(x+sinx)) x lim as x->o+(lnx)
    = 0(-∞)
    = 0
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. May 30, 2013 #2

    haruspex

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    0 × ∞ is indeterminate. When you get that result in a limits question it tells you you need to use a subtler approach.
    What does sin(x) approximate as x tends to zero? Does that allow you to simplify it a little?
    Can you see a variable substitution that would then allow you to lose the sqrt?
     
  4. May 30, 2013 #3
    as sin(x) approaches 0 as x approaches 0.

    That didn't help me as I still resolve to 0*(-∞) but as you say we need a different approach.

    I did try to do" let y = (x+sin(x))^2 " and try to take that approach but I didnt know what to do after as now I am introducing y all of a sudden and dont know how ln x is affected.
     
  5. May 30, 2013 #4

    haruspex

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    No, I was asking for some simpler function that it approximates on its way to 0.
     
  6. May 30, 2013 #5
    It approximates cos(x+ (pi/2)) ?
     
  7. May 30, 2013 #6

    haruspex

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    No, I said simpler. Here's a clue, do you know what the limit of sin(x)/x is as x tends to 0?
     
  8. May 30, 2013 #7
    0/0 ?
     
  9. May 31, 2013 #8

    HallsofIvy

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    Then you need to review (or learn) limits.
     
  10. May 31, 2013 #9

    haruspex

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    0/0 is useless for the same reason that 0 × ∞ is useless. In these situation you get nowhere by taking the limit of each part independently.
    What is the tangent to sin(x) at x = 0?
     
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