Re: Is 1/(x-a)^2 locally Lipschitz continuous?

I understand it can't be globally Lipschitz, because of the
unbounded derivative at the singularity. However, my wife
and i fight about whether it's locally Lipschitz.

I say it is too. She claims foolishly it's not.
Who's the man here?

Isn't it way, way more than just locally Lipschitz?

Close to the x = a you'll be experiencing a rather fast
growth of the function since it's a singularity there.
I don't think it's Lipschitz since i can't find K \in R
such that it still covers for a huge change in vertical
dimension while x and y are very close to eachother.

Did i miss something?

--
Vänligen
Konrad
---------------------------------------------------

Sleep - thing used by ineffective people
as a substitute for coffee

Ambition - a poor excuse for not having
enough sence to be lazy
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