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Delta - epsilon proof

  1. Apr 21, 2004 #1
    Ok well i did problems like this before but now im having trouble with this one for some reason.

    Let [tex]f(x) = \frac{1}{\sqrt{x}}[/tex]. Give a [tex]\delta[/tex] - [tex]\epsilon[/tex] proof that [tex]f(x)[/tex] has a limit as [tex]x \rightarrow 4[/tex].

    So the defn of a limit is
    [tex]\forall \epsilon > 0 \exists \delta > 0[/tex] such that whenever [tex]0 < |x - 4| < \delta[/tex] then [tex]|f(x) - l| < \epsilon[/tex]
    Assuming the limit we are trying to prove is [tex]l[/tex].

    So i know i somehow have to turn [tex]|f(x) - l| < \epsilon[/tex] into something with [tex]x - 4 < ...[/tex] and that will prove that the limit exists. Am i correct? Am i on the right track? can i assume that [tex]l = \frac{1}{2}[/tex] since [tex]\frac{1}{\sqrt{4}}[/tex]?
     
  2. jcsd
  3. Apr 22, 2004 #2

    arildno

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    1. There exists no general technique to find a limit.
    What the d,e-business is about in this case (I think), is to investigate whether an arbitrarily chosen number l is, in fact, the limit value.
    So, it is perfectly legitimate to investigate whether the number 1/2 is the limit value or not.
     
  4. Apr 22, 2004 #3
    Well, I tried to do this, but I sort of got stuck at the end, so if anyone can check & let us know where I've gone wrong and how to complete it correctly, I'd appreciate it.

    I started with the "guess" that the limit is 1/2, and therefore we need to show that
    [tex] \left| \frac{1}{\sqrt{x}} - \frac{1}{2} \right| < \epsilon \: \text{whenever}\: 0 < |x-4| < \delta[/tex]
    If I combine the fractions and require δ < 1, then 3 < x < 5 and √x > 1, so
    [tex] \left| \frac{1}{\sqrt{x}} - \frac{1}{2} \right| = \left| \frac{2 - \sqrt{x}}{2\sqrt(x)}\right| < \left| \frac{2 - \sqrt{x}}{2}\right| < \epsilon [/tex]

    [tex] -\epsilon \;< \; \frac{2 - \sqrt{x}}{2} \;<\; \epsilon [/tex]

    [tex] -2\epsilon \;< \; 2 - \sqrt{x} \;<\; 2\epsilon [/tex]

    [tex] -2\epsilon -2 \;<\; - \sqrt{x} \;<\; 2\epsilon -2 [/tex]

    [tex] 4\epsilon^2 +8\epsilon +4 \;>\; x \;>\; 4 - 8\epsilon + 4\epsilon^2[/tex]

    [tex] 4\epsilon^2 + 8\epsilon \;>\; x-4\; >\; 4\epsilon^2 - 8\epsilon[/tex]

    [tex] 4\epsilon^2 - 8\epsilon \text{ is no good, since for small values of }\epsilon \text { that will be negative,}[/tex]
    [tex]\text{ so does this mean that for any value of }\epsilon \text{ , choosing}[/tex]
    [tex] \delta \;<\; 4\epsilon^2 + 8\epsilon \:\text{satisfies the proof?}[/tex]

    I really don't feel good about this conclusion and I'm getting confused over the directions of the inequalities, so someone please help.

    Edit: well, the more I look at it, the more convinced I am that this is wrong, so if anyone can help, please do.
     
    Last edited: Apr 22, 2004
  5. Apr 22, 2004 #4

    matt grime

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    |2- sqrt(x)| < (2+sqrt(x))(2+sqrt(x)) = |4-x| < d

    that any help?
     
  6. Apr 22, 2004 #5
    No, I still don't see it.
    I assume you meant to write |2- sqrt(x)| < (2-sqrt(x))(2+sqrt(x)) = |4-x| < d

    But we need
    [tex] \left| \frac{2 - \sqrt{x}}{2\sqrt{x}}\right| < \epsilon[/tex]

    whenever
    [tex] \left|2-\sqrt{x}\right| \left|2+\sqrt{x}\right| = \left|4-x\right| < \delta [/tex]

    and I'm still not seeing how to relate ε to δ

    Presumably we need
    [tex]\delta = ({2+\sqrt{x})\epsilon}[/tex]
    but I don't know how to accomplish that.
     
  7. Apr 22, 2004 #6

    Hurkyl

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    I bet you know something smaller than (2 + &radic;x)&epsilon; that doesn't depend on x!
     
    Last edited: Apr 22, 2004
  8. Apr 22, 2004 #7
    I should probably stick to my own homework. :redface:

    That was all wrong. Still working on it.
     
    Last edited: Apr 22, 2004
  9. Apr 22, 2004 #8
    OK, now I let δ = 3ε, and keep the restriction 3<x<5 which is reasonable since we're interested in values of x close to 4.

    So if
    [tex] 0 < |x-4| < \delta[/tex]
    then
    [tex] 0 < |\sqrt{x}-2|\times(\sqrt{x}+2) < 3\epsilon[/tex]
    Now divide by 3
    [tex] 0 < |\sqrt{x}-2|\times \frac{(\sqrt{x}+2)}{3} < \epsilon[/tex]
    And since √x is definitely bigger than 1, we know that
    [tex] \frac{(\sqrt{x}+2)}{3} > 1[/tex]
    so
    [tex] \left| \frac{1}{\sqrt{x}} - \frac{1}{2} \right| = \left| \frac{2 - \sqrt{x}}{2\sqrt{x}}\right| < \left|{2 - \sqrt{x}}\right| < |\sqrt{x} - 2|\times \frac{(\sqrt{x}+2)}{3} < \epsilon [/tex]

    Is that it? Did I spend all this time just to find that? :eek:

    Now it looks right to me. Is anything missing?

    Is there some obvious clue that I should have seen that would have told me to try δ = 3ε?

    Thanks for your help, Matt & Hurkyl. Please let me know if this still needs more work.

    (Edited to correct typos.)
     
    Last edited: Apr 23, 2004
  10. Apr 23, 2004 #9

    Hurkyl

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    Aside from the typo, looks right.

    Incidentally, if you wanted to sew up the loose end, the trick to saying:

    "OK, now I let &delta; = 3&epsilon;, and keep the restriction 3<x<5 which is reasonable since we're interested in values of x close to 4."

    is "let &delta; be the smaller of 3&epsilon; and 1"

    So that way you have both |x-4| < 3&epsilon; and |x-4| < 1
     
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