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The limit of sqrt(x) as x goes to 0

  1. Jun 9, 2006 #1
    My graphing calculator says this is 0, but I don't really know how obvious that answer is. My problem is that it's my understanding that in order for a limit to exist, then the one sided limit from either side of the point in question must converge to the same value. What exactly is is the limit of sqrt(x) from [tex]0^-[/tex]?

    If the square root function was plotted in the x-y plane, and then the complex z plane was plotted perpendicular to this plane, then the square root function would shoot off into the x-z plane for negative values of x. The limit from the negative side of 0 is then clearly toward 0*i, while the limit from the positive side is toward 0. Can the limit be interpreted as existing, using this logic? It's kind of a simple question but I hadn't really thought that I needed to question it before until now.
     
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  3. Jun 9, 2006 #2

    AKG

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    Do you know the exact definition of a limit? Also, 0*i = 0, so what's your point there? Whether you regard the square-root function as a function of the complex numbers (in which case you have to look at a particular branch) or a function of the positive reals, the limit at 0 is 0.
     
  4. Jun 9, 2006 #3
    I couldn't define it if you asked me, to be honest. I know some general things about limits, and I mentioned what I thought proved the existence of a limit, but no, not a definition. I'm aware that 0*i is zero, but the graphical interpretation is what threw me. I couldn't explain to you what it was, but I just needed confirmation on something as stupid. Everyone has those little random stumbling blocks, and this is my one :wink:
     
  5. Jun 9, 2006 #4

    AKG

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    For your knowledge, a real-valued function f with domain D has a limit L at the point a in D iff:

    [tex](\forall \epsilon > 0)(\exists \delta > 0)(\forall x \in D)(|x-a| < \delta \Rightarrow |f(x) - L| < \epsilon)[/tex]

    What this means, roughly, is that for every x IN THE DOMAIN D near a, the value of x is near L. Since negative numbers are not in the domain of the square-root function when its defined on the non-negative reals, the question "how does the square root function approach from the left" is irrelevant.
     
  6. Jun 9, 2006 #5
    One way to look at it.
    f(x) = sqrt(x) is continuous at 0, hence lim x->0 sqrt(x) = sqrt(0) = 0.

    Or using the definition of a limit, we need to show that for all e > 0 there is some d > 0 such that if |x - 0| < d, then |sqrt(x) - 0| < e. Choose d = e^2, then for all e > 0 if |x| < d, then |sqrt(x)| < sqrt(d) = sqrt(e^2) = e and thus lim x->0 sqrt(x) = 0.
     
    Last edited: Jun 9, 2006
  7. Jun 10, 2006 #6

    HallsofIvy

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    Your understanding is not exactly right. Since, as a real valued function, sqrt(x) is not defined for x< 0, only the limit from the right matters. If you are thinking of sqrt(x) as a complex valued function, then there is no problem.
     
  8. Jun 12, 2006 #7

    Alkatran

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    Quick related question: how is the < relation defined on C? Is it based on the length of the hypotenus from 0,0 to your given number (in which case it would only be a partial order)?
     
  9. Jun 12, 2006 #8

    AKG

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    It's not defined on C. It's generally implied that when someone says [itex]\epsilon > 0[/itex], they mean [itex]\epsilon \in \mathbb{R}[/itex].
     
  10. Jun 13, 2006 #9

    HallsofIvy

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    It is not possible to define "<" on the complex numbers in a way that makes C an "ordered field". Of course, the absolute value of a complex number, |z|, is a real number- that's why we always have [itex]|f(z)-f(z_0)|< \epsilon[/itex] and [itex]|z- z_0|< \delta[/itex].
     
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