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Math Help - uniformly continuous

  1. #1
    Junior Member
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    uniformly continuous

    Hello, I need to show that h(x) = 1/(x^2 +1) is uniformly continuous on R

    This is what I have so far:
    Let x,y be in R
    I know |x-y|< delta
    I want to show |f(x)-f(y)|< epsilon taking some delta
    so, |1/(x^2 +1) - 1/(y^2 +1)|
    = |(y^2-x^2)/(x^2 +1)(y^2 +1)|
    = |(x-y)(x+y)/(x^2 +1)(y^2 +1)|
    < |delta(x+y)/(x^2 +1)(y^2 +1)|
    so i could chose an epsilon = (x^2 +1)(y^2 +1)/(x+y)
    so that = epsilon.

    However, to be uniformly continuous, my choice of delta shouldn't depend on x to be uniformly continuous.
    please provide me with some feedback

    thanks
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  2. #2
    Senior Member Tinyboss's Avatar
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    There's a theorem about differentiable functions with bounded derivative. Have you covered it yet?
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  3. #3
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    No, we haven't covered that
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  4. #4
    Senior Member Tinyboss's Avatar
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    The proof is pretty straightforward, and is probably the best way to tackle this. Intuitively, if the graph can only get *this* steep and no steeper, for every x in the reals, can you see why that lets you choose an epsilon that depends only on delta and not on x?
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