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Valid L'Hopitals method for this limit?

  1. Dec 11, 2011 #1
    1. The problem statement, all variables and given/known data

    Find the limit of

    [itex]\frac{x^{2}cos(1/x)}{sin(x)}[/itex] as x tends to 0.

    It doesn't actually say with L'Hopitals rule, but up until now that's what we've been dealing with.


    3. The attempt at a solution

    Wolfram alpha tells me the limit is zero. I know already how to compute the limit of [itex]x^{2}cos(1/x)[/itex] as x tends to 0, and this gives zero.

    By L'Hopitals rule;

    lim(x->0) [itex]\frac{f(x)}{g(x)}[/itex] = lim(x->0) [itex]\frac{f'(x)}{g'(x)}[/itex]


    Can I represent lim(x->0) [itex]\frac{f(x)}{g(x)}[/itex] as lim(x->0) f(x) * lim(x->0) [itex]\frac{1}{g(x)}[/itex]?

    And thus if I show f(x) tends to a value, and 1/g(x) tends to a value, then the whole thing tends to a value? Taking f(x) = [itex]x^{2}cos(1/x)[/itex] and 1/g(x) as [itex]\frac{1}{sin(x)}[/itex], then I can compute the limit of [itex]\frac{1}{sin(x)}[/itex] as the limit of [itex]\frac{e^{x}}{e^{x}sin(x)}[/itex]. Using L'hopitals rule once on this gives me a limit of one. Thus the limit of the whole thing I've wanted is 0/1, so 0.

    Is this correct?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Dec 11, 2011 #2
    Sure you can, it's a property of limits:
    eq0005M.gif
     
  4. Dec 11, 2011 #3

    vela

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    As long as the limits exist, of course.
     
  5. Dec 11, 2011 #4

    Dick

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    I would think of it as (x/sin(x))*(x*cos(1/x)). I'd use l'Hopital for the first factor, if you don't already know it. The second factor isn't all that indeterminant, is it? Use another test on the second factor.
     
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