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Epsilon, delta convergence

  1. Aug 5, 2010 #1
    Hello all,
    My question is as follows:
    f:[1,[tex]\infty[/tex]) is defined by f(x)=[tex]\sqrt{x}[/tex]+2x (1[tex]\leq[/tex]x<[tex]\infty[/tex]) Given [tex]\epsilon[/tex]>0 find [tex]\delta[/tex]>0 such that if |x-y|<[tex]\delta[/tex] then |f(x)-f(y)|<[tex]\epsilon[/tex]

    It seems im being asked to show continuity, and not uniform continuity, so my approach is this, but im not sure it works:
    |f(x)-f(y)|=[tex]\sqrt{x}[/tex]-[tex]\sqrt{y}[/tex] + 2(x-y) after minor manipulation

    since [tex]\sqrt{x}[/tex] function is uniformly continuous it is safe to set [tex]\sqrt{x}[/tex]-[tex]\sqrt{y}[/tex]<[tex]\delta[/tex]/2 and defining |x-y|<[tex]\delta[/tex]/4 we get

    [tex]\sqrt{x}[/tex]-[tex]\sqrt{y}[/tex] + 2(x-y) [tex]\leq[/tex] [tex]\delta[/tex]/2 + 2[tex]\delta[/tex]/4=[tex]\delta[/tex]
    since [tex]\delta[/tex] can equal [tex]\epsilon[/tex] we just need make |x-y|<[tex]\epsilon[/tex]/4

    any help would be appreciated, im not sure i have the right idea here. thank you.
     
  2. jcsd
  3. Aug 5, 2010 #2
    Yes, if you worded the problem correctly, then this will work because all it's asking is to find some [tex] \delta [/tex] that will work. However, just for kicks and giggles, see if you can find the least upper bound for [tex] \delta .[/tex]

    EDIT: On second thought, there's really no need to go that extra step. Nothing really insightful to be gained. Forget it. Your answer's just fine.
     
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