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Math Help - Is the sqrt(x) uniformly continuous on [0,inf)?

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    Is the sqrt(x) uniformly continuous on [0,inf)?

    Is \sqrt{x} uniformly continuous on [0,\infty)? Justify your answer.

    (Note: It's slope at 0 is infinite)
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    yes, it is.

    let's break that set into [0,1]\cup[1,\infty).

    note that the first set is compact, so f is automatically uniformly continuous there, but as for the second set, for each x,y\ge1 is \left| \sqrt{x}-\sqrt{y} \right|=\frac{\left| x-y \right|}{\sqrt{x}+\sqrt{y}}\le \frac{1}{2}\left| x-y \right|, and this satisfies Lipschitz condition, hence we have uniform continuity on the second set, and finally on the whole set.
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    Quote Originally Posted by Krizalid View Post
    yes, it is.

    let's break that set into [0,1]\cup[1,\infty).

    note that the first set is compact, so f is automatically uniformly continuous there, but as for the second set, for each x,y\ge1 is \left| \sqrt{x}-\sqrt{y} \right|=\frac{\left| x-y \right|}{\sqrt{x}+\sqrt{y}}\le \frac{1}{2}\left| x-y \right|, and this satisfies Lipschitz condition, hence we have uniform continuity on the second set, and finally on the whole set.
    You could also note that it's Lipschitz continuous on [1, infinity) since its derivative is bounded there.
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