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Math Help - Calculus Proof

  1. #1
    Newbie Sterwine's Avatar
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    Calculus Proof

    Prove that if there is a number B such that |f(x)| <= B for all x doesnt equal 0, then limit x->0 xf(x) = 0



    Where do I start?
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  2. #2
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    Quote Originally Posted by Sterwine View Post
    Prove that if there is a number B such that |f(x)| <= B for all x doesnt equal 0, then limit x->0 xf(x) = 0



    Where do I start?
    I assume you proved that if g(x) is bounded and \lim_{x \to a}h(x) \to \infty or \lim_{x \to a}h(x) \to 0, then \lim_{x \to a}h(x)g(x) = \lim_{x\to a}h(x)...

    If you have, simply use it and finish the proof. Otherwise, prove it and finish the proof! :P
    Last edited by Defunkt; October 20th 2009 at 04:54 PM. Reason: typo
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  3. #3
    Newbie Sterwine's Avatar
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    Well I dont really understand how you start that but thanks for at least helping
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  4. #4
    Math Engineering Student
    Krizalid's Avatar
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    Quote Originally Posted by Sterwine View Post
    Prove that if there is a number B such that |f(x)| <= B for all x doesnt equal 0, then limit x->0 xf(x) = 0



    Where do I start?
    since \lim_{x\to0}x=0 then for each \epsilon>0 we can let |x|<\frac\epsilon{B}.

    on the other hand, all we need to prove is that |xf(x)|<\epsilon, in efect \left| xf(x) \right|=\left| x \right|\left| f(x) \right|<\frac{\epsilon }{B}\cdot B=\epsilon, and we're done.
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