Re: Proving the existence of a limit



On Fri, 8 Sep 2006, Zdislav V. Kovarik wrote:



On Fri, 8 Sep 2006, Paul Smith wrote:

Even if you exclude that line from the domain, the function takes on
arbitrarily large values in every neighborhood of (0,0).

Thanks, Dave, but how can one prove that?

Paul

Once again, it was V(x,y) = x*y/(y-x)^(1/2) near (0,0).

It is well-defined on the halfplane y>x, and (0,0) is an accumulation
point - so far so good.

I did the arithmetic, and here we go: for every k>0 (to be on the safe
side), the curve given by

y = x + x^4 / (k + (k^2 - x^3))^2

defined for 0 < x < k^(2/3), lies in that halfplane, has (0,0) as an
accumulation point, and the value of the expression V(x,y) along the curve
is up to you to calculate. (Experiment first, then simplify.) So, why does
V(x,y) fail to have a limit, big time?

And: How did I discover the curve???

Cheers, ZVK(Slavek).

.... and Step 4 in Problem Solving: Look back and simplify/improve/modify:

For L>0, what is the limit of V(x,y) as x goes to 0, along the
simple-looking curve

y = x + x^4 / L^2 ?

Cheers, ZVK(Slavek).
.

Relevant Pages

  • Re: Proving the existence of a limit
    ... arbitrarily large values in every neighborhood of. ... It is well-defined on the halfplane y>x, and is an accumulation ... accumulation point, and the value of the expression Valong the curve ...
    (sci.math)
  • Re: Question on Lie Groups
    ... there is a continuous curve from ... > connectedness, but I think for manifolds they are equivalent. ... > If U is a neighborhood of the identity, then cover this curve by left ... I would like to add that I prefer the topological proof already given ...
    (sci.math)
  • Re: Question on Lie Groups
    ... connectedness, but I think for manifolds they are equivalent. ... curve is homeomorphic to the unit interval and hence is compact. ... If U is a neighborhood of the identity, then cover this curve by left ... Since the curve is compact, a finite subcover can be selected. ...
    (sci.math)
  • Re: Question on Lie Groups
    ... there is a continuous curve from ... > connectedness, but I think for manifolds they are equivalent. ... > If U is a neighborhood of the identity, then cover this curve by left ... > Since the curve is compact, a finite subcover can be selected. ...
    (sci.math)
  • Re: Proving the existence of a limit
    ... It is well-defined on the halfplane y>x, and is an accumulation ... accumulation point, and the value of the expression Valong the curve ... How did I discover the curve??? ...
    (sci.math)