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Math Help - Discontinuity...again

  1. #1
    Super Member Showcase_22's Avatar
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    Discontinuity...again

    Is f(x)=[ \sqrt{x} ] where f:[0, \infty) \rightarrow \mathbb{R} is continuous at 0?

    I put false since I can see that the function is not left continuous so it cannot be continuous.

    is this right?
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  2. #2
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    Quote Originally Posted by Showcase_22 View Post
    Is f(x)=[ \sqrt{x} ] where f:[0, \infty) \rightarrow \mathbb{R} is continuous at 0?

    I put false since I can see that the function is not left continuous so it cannot be continuous.

    is this right?
    I assume that you meant simply f(x)= \sqrt{x}. We only require that \lim_{x\rightarrow a} f(x)= f(a) for x in the domain of f in order that f be continuous at a. The domain of \sqrt{x} is all non-negative x. So, no, the fact that \sqrt{x} is not defined for negative x does not prevent it being continuous at x= 0.
    Last edited by Plato; May 31st 2009 at 04:43 AM.
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  3. #3
    Super Member Showcase_22's Avatar
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    I did actually mean f(x)=[\sqrt{x}]

    By your idea the function is continuous at 0 because it's sequentially continuous?
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