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!

The Limit of a Root of a Function is the Root of that Limit (delta-epsilon)

  1. Dec 31, 2011 #1
    1. The problem statement, all variables and given/known data

    Prove:

    If [itex] \lim_{x \rightarrow a} f \left( x \right) = l [/itex], then, if [itex] \sqrt[n]{f \left( x \right)} [/itex] exists, [itex] \lim_{x \rightarrow a} \sqrt[n]{f \left( x \right)} = \sqrt[n]{l} [/itex].

    2. Relevant equations

    Nothing really...

    3. The attempt at a solution

    Okay, so here goes my attempt...

    Proof. We can assume that

    [itex] \forall \varepsilon > 0 [/itex], [itex] \exists \delta_1 > 0 [/itex] : [itex] \forall x [/itex], [itex] 0 < \left| x - a \right| < \delta_1 \Longrightarrow \left| f \left( x \right) - l \right| < \varepsilon [/itex],​

    whereas, we want to show

    [itex] \forall \varepsilon > 0 [/itex], [itex] \exists \delta_2 > 0 [/itex] : [itex] \forall x [/itex], [itex] 0 < \left| x - a \right| < \delta_2 \Longrightarrow \left| \sqrt[n]{f \left( x \right)} - \sqrt[n]{l} \right| < \varepsilon [/itex].​

    But, [itex] \varepsilon > \left| f \left( x \right) - l \right| \geq \left| \sqrt[n]{f \left( x \right)} - \sqrt[n]{l} \right| [/itex]. Therefore, we conclude that given [itex] \varepsilon > 0 [/itex], if, for all [itex] x [/itex], [itex] 0 < \left| x - a \right| < \delta [/itex] for some such [itex] \delta [/itex], then we have both [itex] \left| f \left( x \right) - l \right| < \varepsilon [/itex] and [itex] \left| \sqrt[n]{f \left( x \right)} - \sqrt[n]{l} \right| < \varepsilon [/itex].

    QED

    So, I guess the first remark would be if my method was even right, where the second would be if my inequality is true (I can't think of a counterexample).

    Anyway, much appreciation for any help.
     
  2. jcsd
  3. Dec 31, 2011 #2

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    For n=2, f(x) = 1/2 and l = 0 is a counterexample.
     
  4. Jan 1, 2012 #3
    Oh, drat. Then, I am at a loss on what to do.

    Hmm... I'm likely misunderstanding, but if [itex] f \left( x \right) = 1/2 [/itex], then wouldn't [itex] l = \lim_{x \rightarrow a} 1/2 = 1/2 [/itex] for any [itex] a [/itex]? That is, I don't understand how the limit [itex] l [/itex] could be [itex] 0 [/itex].

    ---

    Much appreciation
     
  5. Jan 1, 2012 #4

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Sorry, what I wrote wasn't very clear. I meant if f=1/2 for some value of x near a, then the implication your proof relied on doesn't hold. I didn't mean that f was a constant function.
     
  6. Jan 1, 2012 #5
    [itex] |\sqrt{f(x)} - \sqrt{L}| = \frac{f(x) - L}{|\sqrt{f(x)} + \sqrt{L}|}[/itex].

    The denominator of the right side of the equation can never be zero unless f = 0 in some neighborhood of x= a and L = 0 (in which case the proposition you're trying to prove is trivial). Why?

    Once you answer that, let x approach a and see what the limit of the right side of the equation is.
     
  7. Jan 4, 2012 #6
    Alright... I've been trying to understand these hints, but I unfortunately don't seem to get anywhere. However, I realized that I don't need this proof as much as I thought, so I am just going to put this on the 'back burner' (if I got that right) for now. Thanks for the help!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: The Limit of a Root of a Function is the Root of that Limit (delta-epsilon)
Loading...