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Prove continuity.

  1. Dec 26, 2013 #1
    1. The problem statement, all variables and given/known data
    Prove that [itex]\sqrt{x}[/itex] is continuous in R+ by using the epsilon-delta definition.


    2. Relevant equations
    A function f from R to R is continuous at a point a [itex]\in[/itex] R if :

    Given ε> 0 there exists δ > 0 such that if |a - x| < δ then |f(a) - f(x)| < ε



    3. The attempt at a solution


    So I have to take an ε which is greater than 0 and prove that there exists a δ such that if the absolute value of (a - x) is smaller than that delta then the absolute values of the function values of a and x are smaller than ε.

    I know I have to pick an ε which is greater than 0 but how do I know what value to pick for ε? 1,2,...n?
     
  2. jcsd
  3. Dec 26, 2013 #2

    Mark44

    Staff: Mentor

    You don't get to pick the ε. The whole δ - ε thing should be viewed as a dialog between you (who are trying to prove that a certain limit exists) and an acquaintance who is skeptical of the process. The other person gives you an ε value, which by the way is usually small and close to zero. You respond by finding a number δ so that when x is within δ of a, then √x is within ε units of √a.

    If your skeptical friend is not satisfied, he will say something like, "Well it works for that ε. How about if ε is smaller?" You respond by finding a different δ, and show your friend that when x is within δ of a, then √x is again within ε units of √a.

    The process continues until your skeptical friend realizes that no matter how small an ε he gives you, you are able to come up with a δ that works, and the limit is established.
     
  4. Dec 26, 2013 #3

    Thank you. This was very helpful intuive-wise! But how do I do this formally? Do I write δ in terms of ε?
    I'm sorry but I'm really stuck here.
     
  5. Dec 26, 2013 #4

    Mark44

    Staff: Mentor

    Yes. Also, you need to work in the actual function, not f(x).

    Start with the inequality |√x - √a| < ε and work backwards to |x - a| < <some expression>. That <some expression> will be your delta.
     
  6. Dec 26, 2013 #5

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    At any point, a, you want to prove that given some [itex]\epsilon> 0[/itex], there exist [itex]\delta> 0[/itex] such that "if [itex]|x- a|< \delta[/itex] then [itex]|f(x)- f(a)|< \epsilon[/itex].

    So start with what you want to get: [itex]|f(x)- f(a)|= |\sqrt{x}- \sqrt{a}|< \epsilon[/itex] and try to manipulate that to get "[itex]|x- a|< [/itex] some number".

    I recommend you start by separating this into two cases: x> a and x< a. Square root is an increasing function so that if x> a then [itex]\sqrt{x}> \sqrt{a}[/itex] and if x< a then [itex]\sqrt{x}< \sqrt{a}[/itex] so you can eliminate the absolute values.
     
  7. Dec 26, 2013 #6

    |√x - √a| < ε


    Sorry, accidentally clicked on post. I'm working on it :p
     
  8. Dec 26, 2013 #7
    Ok, so I've, actually it was you guys, come up with this so far :

    |√x - √a| < ε

    ⇔ |√x - √a| = |√x - √a| . [itex]\frac{|\sqrt{x}+ \sqrt{a}|}{|\sqrt{x}+ \sqrt{a}|}[/itex]

    ⇔ |√x - √a| = [itex]\frac{|x-a|}{|\sqrt{x}+ \sqrt{a}|}[/itex]

    ⇔ [itex]\frac{|x-a|}{|\sqrt{x}+ \sqrt{a}|}[/itex] < ε

    ⇔ |x-a| < ε . [itex]|\sqrt{x}+ \sqrt{a}|[/itex]

    Is this correct so far?
     
  9. Dec 26, 2013 #8

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    You've got the key equality, which is to relate √x - √a to x - a.

    |√x - √a| = |x−a|/|√x +√a|

    I would do this first for a = 0. I.e. prove √x is continuous at 0.

    Then prove it for a > 1. If a > 1, then, |√x - √a| is smaller than |x−a|. So, it should be easy to find δ. But, if a < 1, then |√x - √a| could be larger than |x−a|. That's the tricky bit.

    So, if you want to take it step by step, you could prove it for a >= 1. And, then finally prove it for a < 1. This might be easier until you get used to ε-δ.

    For a < 1, there's a trick you'll need that is used a lot in ε-δ. It's not easy to spot first time you come across it.
     
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