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Homework Help: L'Hopital! problem help

  1. Dec 8, 2009 #1
    Estimate the limit using L'Hopital's rule:

    1. limit of x is going to infinity of xtan(1/x)

    Okay, I have no idea how to do #1, but i know the indetermiate form is infinity*0.

    2. limit as x tends to 0+ of (lnx - ln sinx)

    For #2 the indeterminate form is infinity-infinity, so i took the derivative of the function and got:

    (1/x) - (cosx/sinx) which yields the same indeterminate form of infinity - infinity.

    So my question is, should i get both with a common denominator, or should i derive (1/x) - (cosx/sinx) again and plug 0 to see if it works?

    Any suggestions? For both 1 or 2?
     
  2. jcsd
  3. Dec 8, 2009 #2

    Mark44

    Staff: Mentor

    Re: L'Hopital!

    1. x*tan(1/x) = [tan(1/x)]/[1/x]
    If you let u = 1/x, then the above is tan(u)/u which has a known limit as u -->0+.

    2. lnx - ln sinx = ln(x/sin x)

    lim ln(x/sin x) = ln(lim (x/sin x)), provided that the limit exists.
     
  4. Dec 8, 2009 #3
    Re: L'Hopital!

    Mark, tan(u)/u is going to be 0/0 which is an indeterminate form?
     
  5. Dec 8, 2009 #4

    Mark44

    Staff: Mentor

    Re: L'Hopital!

    Do you have a question about what I said?
     
  6. Dec 8, 2009 #5
    Re: L'Hopital!

    Well, tan(u)/u is 0/0 which is an indeterminate form. Do i take the derivative of tan(u)/u then substitute "u" back into the equation? I'm confused on what you did for #2?
     
  7. Dec 8, 2009 #6
    Re: L'Hopital!

    for 1, in my opinion, i don't like u substitution so I just evaluate the limit without substituting.
    I would do what Mark44 said and form tan(1/x)/(1/x)
    as you should see, you'll get 0/0 for which you can take the derivatives and go from there
     
  8. Dec 8, 2009 #7
    Re: L'Hopital!

    for number 2, the properties of a logarithm were applied and since ln is a continuous function, you can take the limit of the inside while keeping ln on the outside for a little
     
  9. Dec 8, 2009 #8
    Re: L'Hopital!

    Ok so, dy/dx = [sec2(1/x)/x2]/(1/x2)
    Then what?

    Any help for #2?
     
  10. Dec 8, 2009 #9
    Re: L'Hopital!

    So, for #2 i can do:

    d/dx ln(lim (x/sin x)) = ln (1/cosx) ?
    plug in 0 and get ln (1/1) = 0?
     
  11. Dec 8, 2009 #10
    Re: L'Hopital!

    for number 2, again do what Mark44 suggested without any substitution(my opinion) if the need arises.
    So, ln(x) - ln (sinx) becomes ln(x/sinx)
    (remember that property of logarithms)(if not, it would be best to look over those again)

    then, you take the limit as x approaches 0 from the right of ln(x/sinx).
    since ln is a continuous function on its entire domain(looking at the graph of ln x should answer any doubts), you can first take the limit as x approaches 0+ of x/sinx, then when you get an answer put it in the ln function like so: ln(answer) = true answer ( can't forget about the ln you took out)
     
  12. Dec 8, 2009 #11
    Re: L'Hopital!

    are you sure 1/1 = 0? and don't forget about the ln you took out to apply the limit
     
  13. Dec 8, 2009 #12
    Re: L'Hopital!

    i said the ln(1/1) = 0

    so if i were to use the ln(x/sinx) function and find the limit as it tends to 0+, i would just get the ln(0/0); so, the answer is just infinity?
     
  14. Dec 8, 2009 #13
    Re: L'Hopital!

    good question
    i would say no because you have an indeterminate form, which wouldn't be infinity
    you need to resolve the problem of the indeterminate form to evaluate the limit
    0/0 is definitely not infinity
     
  15. Dec 8, 2009 #14
    Re: L'Hopital!

    Oh, i see.
    Okay, i think i understand the situation now, thanks!
     
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