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Math Help - L'Hospital's Rule to show non-existence of limits

  1. #1
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    L'Hospital's Rule to show non-existence of limits

    Let f(x)=x^2\sin{1/x} for 0<x\leq 1 and f(0)=0 , and let g(x)=x^2 for x\in [0,1] .
    Then both f and g are differentiable on [0,1] and g(x)>0 for x\neq 0 .
    Show that \lim_{x \to 0} f(x) = 0 = \lim_{x \to 0} g(x) and that \lim_{x \to 0} f(x)/g(x) does not exist.
    Last edited by frenchguy87; April 7th 2010 at 03:53 PM.
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  2. #2
    Super Member Failure's Avatar
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    Quote Originally Posted by frenchguy87 View Post
    Let f(x)=x^2\sin{1/x} for 0<x\leq 1 and f(0)=0 , and let g(x)=x^2 for x\in [0,1] .
    Then both f and g are differentiable on [0,1] and g(x)>0 for x\neq 0 .
    Show that \lim_{x \to 0} f(x) = 0 = \lim_{x \to 0} g(x) and that \lim_{x \to 0} f(x)/g(x) does [?] exist.
    But surely, there must be a typo in the statement of your problem: \lim_{x\to 0+}f(x)/g(x) does not exist, since it boils down to \lim_{x\to 0+}sin(1/x).
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  3. #3
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    fixed!
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  4. #4
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    does anyone know?
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  5. #5
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    Quote Originally Posted by frenchguy87 View Post
    does anyone know?
    Where are you having problems? This is rather straight forward.
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  6. #6
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    Never mind I figured it out earlier.
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