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Thread: Question about applying L'Hospitals rule

  1. #1
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    Question about applying L'Hospitals rule

    Hi,

    Say I have a limit that, on a first direct evaluation, gives (0 x infinity)/0. Can I still use L'Hospitals rule? The specific limit I've been given is (x^2(sin(1/x)))/sinx as x approaches 0. It seems that I could divide the numerator by 1/sin(1/x)), which would give 0/0/0 which doesn't seem to help. Is it even possible?

    Thanks!
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  2. #2
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    Re: Question about applying L'Hospitals rule

    Question about applying L'Hospitals rule-31-jan-14.png
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  3. #3
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    Re: Question about applying L'Hospitals rule

    Hi,

    x^2\sin(1/x) does approach 0 as x approaches 0. One way to see this:

    |\sin(z)|\leq1 for any z, so 0\leq|x^2\sin(1/x)\leq|x^2| for x not 0. The squeeze theorem says the limit is 0.

    So you can try l'hopital's rule by differentiating both numerator and denominator. However, when you do this you get a limit which does not exist. So to find the limit, you must try something else. You should check on a possible limit by graphing the function and zooming in towards the origin.

    Post again if you still can't figure it out.
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  4. #4
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    Re: Question about applying L'Hospitals rule

    Johng, I had to think about it but your explanation makes good sense, thank you.

    Ibdutt, now I know how to solve using l'Hospitals rule when I get limits like (0 x infinity) / 0. Thanks!
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