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Prove existence of a limit

  1. Jan 13, 2013 #1
    1. The problem statement, all variables and given/known data

    given
    ##A \subset \mathbb{R}##

    ##f:A \subset \mathbb{R} \to \mathbb{R}^+##

    considering the function g such that:

    ##g(x):=\sqrt{f(x)} x \in A## with ##x_0## limit point in A.
    Prove that if ##\displaystyle \lim_{x \to x_0} f(x)## exists, then ##\displaystyle \lim_{x \to x_0} g(x)=\sqrt{\displaystyle \lim_{x \to x_0} f(x)}##

    3. The attempt at a solution

    Being ## \displaystyle \lim_{x \to x_0}=L##
    By definition of limit:
    ##\forall \epsilon>0 \exists \delta >0: \forall x: 0<|x-x_0|<\delta \Rightarrow |f(x)-L|<\epsilon##

    Let: ##\epsilon=\epsilon |\sqrt{f(x)}+\sqrt L|##

    ##|f(x)-L|<\epsilon |\sqrt{f(x)}+\sqrt L| \Rightarrow |\frac{f(x)-L}{\sqrt{f(x)}+\sqrt L}|< \epsilon## ##\Rightarrow |\sqrt{f(x)}+\sqrt L}|<\epsilon##

    is it sufficient? thank you in advance
     
  2. jcsd
  3. Jan 13, 2013 #2

    pasmith

    User Avatar
    Homework Helper

    Sadly this is not sufficient.

    Your aim is to prove:
    [tex](\forall \epsilon > 0)(\exists \delta > 0)(\forall x \in A)(|x - x_0| < \delta \Rightarrow |g(x) - \sqrt L|< \epsilon).[/tex]

    There are two ways to do this. The easiest is to use the fact that the square root function is continuous, which tells you that for all [itex]\epsilon > 0[/itex] there exists [itex]\eta > 0[/itex] such that for all [itex]f(x) \geq 0[/itex], if [itex]|f(x) - L| < \eta[/itex] then [itex]|g(x) - \sqrt L | < \epsilon[/itex].

    The second way is to let [itex]\epsilon > 0[/itex] be arbitrary, start from
    [tex]|g(x) - \sqrt L| = \left|\sqrt{f(x)} - \sqrt L\right|< \epsilon[/tex]
    and multiply by [itex]\sqrt{f(x)} + \sqrt L[/itex] to get
    [tex]
    |f(x) - L| - \epsilon\left(\sqrt{f(x)} + \sqrt L\right) < 0
    [/tex]
    In each of the cases [itex]f(x) > L[/itex] and [itex]f(x) < L[/itex] the left hand side is a quadratic in [itex]\sqrt{f(x)}[/itex], so you can derive conditions on [itex]\sqrt{f(x)}[/itex] for the inequality to hold. You should be able to rearrange these to get an upper bound on [itex]|f(x) - L|[/itex] in terms of [itex]\epsilon[/itex] and [itex]L[/itex].

    (What you're actually doing here is proving that square root is continuous and using a slightly cumbersome notation in order to do so.)
     
  4. Jan 13, 2013 #3

    SammyS

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    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    I fixed some obvious typos (in red).

    From the above, you know that given any ε, you can find a δ that works for f(x).

    Now, what you need to show is that for any (arbitrary) ε>0, you can find a δ which works for g(x).

    You don't get to choose an ε, which is what you have below. Also, when finding a δ, it won't depend on x. I can possibly depend on x0, but not on x .
     
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