1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    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


    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


    User Avatar
    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


    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted