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Math Help - More uniform continuity!!

  1. #1
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    More uniform continuity!!

    iF g(x)=\sqrt{x} for x\in [0,1] , how would you show that there doesn't exist a constant K such that |g(x)|\leq K|x| , for x\in [0,1]
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  2. #2
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by CrazyCat87 View Post
    iF g(x)=\sqrt{x} for x\in [0,1] , how would you show that there doesn't exist a constant K such that |g(x)|\leq K|x| , for x\in [0,1]
     |g(x)|\leq K|x| \implies \sqrt{x} \leq Kx \implies \frac{\sqrt{x}}{x} \leq K \implies \frac{1}{\sqrt{x}} \leq K \; \forall x\in [0,1] .

    Hence  \lim_{x\to 0} \frac{1}{\sqrt{x}} \leq K , but  \lim_{x\to 0} \frac{1}{\sqrt{x}} = \infty .

    Therefore  K can't exist for all  x in this domain.
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  3. #3
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    Could you conclude this is a Lipschitz function?
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  4. #4
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    no it's not Lipschitz.
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  5. #5
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    why??
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