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limits of rational functions and C^inf
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echabbadon  
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 More options Oct 28 2009, 11:59 pm
Newsgroups: sci.math
From: echabbadon <kypri...@indiana.edu>
Date: Wed, 28 Oct 2009 19:59:43 EDT
Local: Wed, Oct 28 2009 11:59 pm
Subject: limits of rational functions and C^inf
Ok, i need some advice on this one:

if
f(x)=
{0,                      if x=0  
p(x)/q(x) *(e^{-1/(x^2)} if x neq 0

where p and q are poly's, defined off the nonzero roots of q.  Prove lim x->0 f(x)=0 and f(x) is infinetely differentiable.

My solution starts like this: lim p(x)/q(x) *(e^{-1/(x^2)}=lim p(x)/q(x) * lim e^{-1/(x^2)} if both limits exist and are finite.   By L'hopitals rule
lim p(x)/q(x)=lim p'(x)/q'(x)... and so on.  If deg(p(x))>deg(q(x)) then this chain equals lim p^(n)(x)/1 which is just some constant. and similiary if deg(p(x))<deg(q(x)).  Thus since lim e^{-1/(x^2)=0 we have it.

But for showing its C^inf im not sure what to do...


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echabbadon  
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 More options Oct 29 2009, 1:43 am
Newsgroups: sci.math
From: echabbadon <kypri...@indiana.edu>
Date: Wed, 28 Oct 2009 21:43:26 EDT
Local: Thurs, Oct 29 2009 1:43 am
Subject: Re: limits of rational functions and C^inf
Please I need help with this one...im at a standstill

When I start taking derivatives of this there are all things that I think are differentiable but I cannot prove it, and it becomes messy really quickly..


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Robert Israel  
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 More options Oct 29 2009, 4:18 am
Newsgroups: sci.math
From: Robert Israel <isr...@math.MyUniversitysInitials.ca>
Date: Wed, 28 Oct 2009 23:18:43 -0500
Local: Thurs, Oct 29 2009 4:18 am
Subject: Re: limits of rational functions and C^inf

Hint 1: write this as g(x) e^(-1/x^2) x^m where m is an integer (may be
positive or negative) and g(x) is a rational function with g(0) <> 0.
Show that the limit as x -> 0 is 0.  

Hint 2: express f' in a form similar to that of f.

Hint 3: mathematical induction.
--
Robert Israel              isr...@math.MyUniversitysInitials.ca
Department of Mathematics        http://www.math.ubc.ca/~israel
University of British Columbia            Vancouver, BC, Canada


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