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A delta-epsilon proof.

  1. Jan 18, 2008 #1
    I'm just kind of confused on how to handle the absolute value along with the inequality. Bad algebra skills. :(

    0 < |x - x_0| < [tex]\delta[/tex] [tex]\Rightarrow[/tex] | f (x) - L | < [tex]\epsilon[/tex]

    I'm given: f(x) = 2[tex]\sqrt{x + 1}[/tex], x[tex]_{0}[/tex] = 3, L = 4, [tex]\epsilon[/tex] = 0.2

    The attempt at a solution

    Nevermind, I tried latex and it messed everything up. Sorry.
     
    Last edited: Jan 18, 2008
  2. jcsd
  3. Jan 19, 2008 #2

    arildno

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    Hi, there!

    It is best to forget about the epsilon at first, and focus on bounding the difference in terms of delta.

    So, we are to bound:
    [tex]|\sqrt{2x+1}-\sqrt{2x_{0}+1}|=|\frac{2x+1-(2x_{0}+1)}{\sqrt{2x+1}+\sqrt{2x_{0}+1}}|=2\frac{|x-x_{0}|}{|\sqrt{2x+1}+\sqrt{2x_{0}+1}|}<2\frac{\delta}{\sqrt{2x_{0}+1}}[/tex]

    You may use this to determine a delta that surely will work for som particular epsilon, say:
    [tex]\delta=\frac{\epsilon}{2}\sqrt{2x_{0}+1}[/tex]

    Edit:
    Seems I used the wrong function, but the technique is similar for your case. Try it out.
     
    Last edited: Jan 19, 2008
  4. Jan 19, 2008 #3
    Why did you do 2x_0 + 1?
     
  5. Jan 19, 2008 #4

    arildno

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    Did you see my edit?

    Follow a similar procedure with |2*sqrt(x+1)-2*sqrt(x0+1)| instead.

    Don't bother to use digit-written numbers(like using 3 instead of x0) before the end.
     
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