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Need help proving a limit using epsilon delta definition.

  1. Sep 24, 2011 #1
    Need help proving lim(x)->(a) sqrt(x)=sqrt(a) using epsilon delta definition.

    1. The problem statement, all variables and given/known data

    Prove that the limit of [itex]\sqrt{x}[/itex] is [itex]\sqrt{a}[/itex] as x approaches a
    if a>0

    2. Relevant equations


    in words

    By the epsilon delta definition we know that the distance between f(x) and the limit of the function will be less than "epsilon" if x is within a certain distance "delta" of a.


    in equation form

    0<|x-a|<delta
    |f(x)-L|<epsilon

    3. The attempt at a solution

    ok i got messed up with the latex and i don't know how to fix it.

    But what I did was multiplied |sqrt(x)-sqrt(a)| by |sqrt(x)+sqrt(a)| over itself to get the top in a form similar to delta. So that comes out to |x-a|/|sqrt(x)+sqrt(a)| is less than epsilon.

    so we have |x-a| in both the epsilon and delta inequalities, so is it safe to write delta in terms of epsilon? I don't know what to do now...
    ???




    i'm stuck
     
    Last edited: Sep 24, 2011
  2. jcsd
  3. Sep 24, 2011 #2
    ok i got messed up with the latex and i don't know how to fix it.

    But what I did was multiplied |sqrt(x)-sqrt(a)| by |sqrt(x)+sqrt(a)| over itself to get the top in a form similar to delta. So that comes out to |x-a|/|sqrt(x)+sqrt(a)| is less than epsilon.

    so we have |x-a| in both the epsilon and delta inequalities, so is it safe to write delta in terms of epsilon? I don't know what to do now...
     
  4. Sep 24, 2011 #3
    Prove that the limit of √x as x approaches a is √a
    if a>0


    please help...
     
  5. Sep 24, 2011 #4

    SammyS

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    Let's fix the Latex problem.
    I fixed the following, too.
     
  6. Sep 24, 2011 #5
    Thank you so much SammyS!

    I have made further progress since I last posted and I may be coming to a solution. I will post what I have soon and hopefully I can get the latex right this time.
     
  7. Sep 25, 2011 #6
    Ok, I got stuck again. Here is what I have...

    0<|x-a|< delta
    |[itex]\sqrt{x}[/itex]-[itex]\sqrt{a}[/itex]| < epsilon

    multiply |[itex]\sqrt{x}[/itex]-[itex]\sqrt{a}[/itex]|
    by
    |[itex]\sqrt{x}[/itex]+[itex]\sqrt{a}[/itex]| over itself
    to get
    |x-a|/[itex]\sqrt{x}[/itex]+[itex]\sqrt{a}[/itex] < epsilon

    Now we need to find a constant C such that
    |x-a|/[itex]\sqrt{x}[/itex]+[itex]\sqrt{a}[/itex] < |x-a|/C < epsilon

    Since we want |x-a|, (AKA "delta") to be a pretty small number, we will choose it to be <1

    in order for |x-a|/C to be > than |x-a|/[itex]\sqrt{x}[/itex]+[itex]\sqrt{a}[/itex], C must be smaller than |[itex]\sqrt{x}[/itex]+[itex]\sqrt{a}[/itex]|

    for the next step we will set up the inequality
    |x-a|< 1
    and try to rearrange it in a way where we can discover what the constant C (which is < than |[itex]\sqrt{x}[/itex]+[itex]\sqrt{a}[/itex]|) could be set equal to to assist in solving the problem.

    solving |x-a|< 1 in terms of x gives us
    x < a[itex]\pm[/itex]1

    so since we can set up this inequality
    C < |[itex]\sqrt{x}[/itex]+[itex]\sqrt{a}[/itex]| < |sqrt{a[itex]\pm[/itex]1}+sqrt{a}




    but now i am stuck. Where have I gone wrong?
     
  8. Sep 25, 2011 #7

    HallsofIvy

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    You want [itex]|\sqrt{x}- \sqrt{a}|<\epsilon[/itex]. Do NOT "multiply by [itex]\sqrt{x}+ \sqrt{a}[/itex] over itself". Instead, multiply both sides by [itex]\sqrt{x}+ \sqrt{a}[/itex] (which is obviously positive) to get [itex]|x- a|< \epsilon(\sqrt{x}+ \sqrt{a})[/itex]. Since [itex]\sqrt{x}[/itex] is positive, [itex]\sqrt{a}< \sqrt{x}+ \sqrt{a}[/itex] and [itex]\epsilon\sqrt{a}< \epsilon(\sqrt{x}+ \sqrt{a})[/itex]. You can take [itex]\delta= \epsilon\sqrt{a}[/itex]. That way, if [itex]|x- a|< \delta= \epsilon\sqrt{a}[/itex] you must also have [itex]|x- a|< \epsilon(\sqrt{x}+\sqrt{a})[/itex].
     
  9. Sep 25, 2011 #8
    Thanks HallsofIvy. Although I can see why every one of your statements is true, I don't understand how it proves the limit as x approaches a of [itex]\sqrt{x}[/itex] is [itex]\sqrt{a}[/itex].

    I just don't understand why you did any of the things that you did. :(
    Could you (or someone else) maybe help me understand what we were trying to accomplish? I know the problem wants us to prove a limit using the epsilon delta definition of a limit, but I don't understand what that entails, and I certainly wouldn't have known what steps to take if you hadn't shown me...




    Edit: Are we able to take [itex]\delta= \epsilon\sqrt{a}[/itex] because we know that [itex]|x- a|< \delta[/itex] and [itex]|x- a|< \epsilon(\sqrt{x}+ \sqrt{a})[/itex], and [itex]\epsilon\sqrt{a}< \epsilon(\sqrt{x}+ \sqrt{a})[/itex], so since |x-a| is less than [itex]\delta[/itex] and |x-a| and [itex]\epsilon\sqrt{a}[/itex] are both less than [itex]\epsilon(\sqrt{x}+ \sqrt{a})[/itex], we are able to set [itex]\delta[/itex] equal to [itex]\epsilon\sqrt{a}[/itex]? But if [itex]\delta[/itex] = [itex]\epsilon\sqrt{a}[/itex], wouldn't that mean that [itex]\epsilon\sqrt{a}[/itex] > |x-a|? And how can we know that?
     
    Last edited: Sep 25, 2011
  10. Sep 25, 2011 #9

    HallsofIvy

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    To prove that "[itex]\lim_{x\to a}f(x)= L[/itex]" means to show that, given any [itex]\epsilon> 0[/itex] there exist [itex]\delta> 0[/itex] such that if [itex]|x- a|<\delta[/itex] then [itex]|f(x)- L|< \epsilon[/itex].

    The point of choosing [itex]\delta= \epsilon\sqrt{a}[/itex] is that you can then work backwards from the way I showed before:
    If [itex]|x- a|< \epsilon\sqrt{a}< \epsilon (\sqrt{x}+\sqrt{a})[/itex]
    then
    [tex]\frac{|x- a|}{\sqrt{x}+ \sqrt{a}}< \epsilon[/tex]
    [tex]\frac{|(\sqrt{x}-\sqrt{a})(\sqrt{x}+\sqrt{a})|}{\sqrt{x}+ \sqrt{a}}< \epsilon[/tex]
    [tex]|\sqrt{x}- \sqrt{a}|< \epsilon[/tex]
     
  11. Sep 25, 2011 #10
    Okay that makes sense too. But where do we go from there to prove that the limit of √x is √a as x approaches a if a>0?

    Can I say: If we want to be within [tex]|\sqrt{x}- \sqrt{a}|[/tex] of the limit of √x, then we must pick an x value that is within |x-a| of a (AKA delta). So the delta we choose will be equal to [itex]\epsilon[/itex] *[tex]|\sqrt{x}- \sqrt{a}|[/tex]...

    now what?
     
    Last edited: Sep 25, 2011
  12. Sep 25, 2011 #11

    SammyS

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    What HallsofIvy gave you in post #7 is what is often done on "scratch-paper", working backwards from |f(x) - L| < ε to |x-a| < δε , where δε is some expression for δ which is (usually) based on ε .

    In final version of the proof you reverse that backwards argument and clean up any rough spots.
    In other words: Using the δε that you found, start with 0 < |x - a| < δε. Then reversing your backwards "scratch-paper" argument, eventually you should arrive at |f(x) - L| < ε .

    In your case, HallsofIvy showed you that δε = ε∙√a should give you your desired result.​
     
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