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Finding modulus of continuity

  1. Apr 28, 2009 #1
    Say f(x) = x^2 - 1 and I'm trying to prove that f is continuous, then I was told I CANNOT do this:

    [tex] |x^2 - x_0^2| = |x-x_0||x+x_0| < \delta|x+x_0| = \epsilon [/tex]

    because then our epsilon is relying on an x value. I was told I could restrict the x values to a closed neighborhood about the point x = x_0, and let epsilon rely on the end points, but I cannot let it rely on any x values like in the example shown.

    Is this correct? And if it is, why is it?
     
  2. jcsd
  3. Apr 28, 2009 #2

    HallsofIvy

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    Yes, it's true. It's true because [itex]\epsilon[/itex] is supposed to be a constant and [itex]\delta|x+ x_0|[/itex] is not a constant. so "[itex]\delta|x+ x_0|= \epsilon[/itex] makes no sense.

    What you can do is this: Since we only need look at x close to x0, assume |x- x0|< 1. The -1< x- x0< 1 so, adding 2x0 to each part, -1+ 2x0< x+ x_0< 1+ 2x0. The larger of those two numbers, in absolute value, is 1+ 2x0 so [itex]\delta|x+ x_0|< \delta(1+ 2x_0[/itex]. We must choose [itex]\delta[/itex] such that [itex]\delta(1+ 2x_0)< \epsilon[/itex]. Actually, because of the requirement that |x- x0|< 1, we must choose [itex]\delta[/itex] to be the smaller of that number and 1.
     
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