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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


    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].


    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


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    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


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    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!
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