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Homework Help: Limit of the form 0^infinity+infinity^0

  1. Dec 8, 2009 #1
    1. The problem statement, all variables and given/known data
    What is the value of
    lim {(1/x)sinx+(sinx)1/x} ,x>0 ???
    x->0


    2. Relevant equations
    ??


    3. The attempt at a solution
    lim {(1/x)sinx+(sinx)1/x}
    x->0
    =lim (1/x)sinx+lim (sinx)1/x
    ..x->0............x->0

    Let A=lim (1/x)sinx
    .........x->0
    and B=lim (sinx)1/x
    .........x->0

    logA=lim (sinx)log(1/x)=0
    ........x->0
    =>A=e0=1

    logB=lim (1/x)log(sinx)=infinity
    ........x->0
    =>B=infinity

    The problem is that the answer is 0.
     
    Last edited: Dec 9, 2009
  2. jcsd
  3. Dec 8, 2009 #2

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    I understand that x goes to 0 in the limit, but how ? [itex] \nearrow 0 [/itex] or [itex] \searrow 0[/itex] ?
     
  4. Dec 8, 2009 #3

    Mark44

    Staff: Mentor

    Split it up into two limits and work with them. If both limits exist, then their sum will be the limit you want.

    Since both expressions involve exponents, you are most likely going to have to take the log, which will probably get you something you can use L'Hopital's Rule on.
     
  5. Dec 9, 2009 #4
    I discussed this problem with someone..
    He says,
    lim (sinx)1/x=1
    x->0
    So, according to him answer must be 2.

    I worked it out again and got the same answer, i.e., infinity...
    How is it happening...??
    Please, show the steps if you have got the answer 2, 1 or 0.
    These were the options given to me...
    However, if you feel that the correct answer is not among the options please tell with explanation and steps.
     
  6. Dec 9, 2009 #5

    Mark44

    Staff: Mentor

    I believe the limit in your first post is 1, but I can't justify it just yet. (sin x)^(1/x) -->0 as x -->0+, and (1/x)^(sin x) --> 1 as x --> 0+. I'll see if I can provide some justification for this result.
     
  7. Dec 9, 2009 #6

    Mark44

    Staff: Mentor

    As x approaches 0 from above. (The OP said that x > 0.)
     
  8. Dec 9, 2009 #7

    Mark44

    Staff: Mentor

    Let y1 = (1/x)sin x
    Then ln y1 = ln[(1/x)sin x] = sinx*ln(1/x) = sinx*(-lnx) = -lnx/cscx
    In the next step I use L'Hopital's Rule, since we have an indeterminate form of [infinity/infinity]. Also, in the following steps, lim means limit as x -->0+
    lim ln[ ln y1] = lim [-lnx/cscx]
    = lim [(-1/x)/(-csc x * cot x)] = lim[(1/x)/(csc x * cot x)]
    = lim[(1/x)/(1/sinx * cos x/sinx)] = lim [sin^2(x)/(x cos x)]
    = lim [(sin(x)/x * tan(x)] = lim[sin(x)/x]*lim tan(x) = 1*0 = 0.

    Since lim ln y1 = 0, ln lim y1 = 0, hence lim y1 = 1.

    The other limit is easier.
    Let y2 = sinx1/x = (eln(sinx))1/x = e(1/x)ln(sin x)

    Before taking the limit, let's look at what the factors in the exponent on e are doing as x --> 0+.

    1/x --> + infinity.
    sin(x) --> 0+, so ln(sin(x)) --> -infinity.
    The product of 1/x and ln(sin(x)) --> -infinity.
    Since the exponent on e is approaching -infinity, then e to that power --> 0.
    Hence lim y2 = lim e(1/x)ln(sin x) = 0.

    Therefore lim (y1 + y2) = 1 + 0 = 1.
     
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