Date: Mar 1, 2013 6:22 PM
Author: Stuart M Newberger
Subject: Re: Differentiability
On Thursday, February 28, 2013 8:33:58 PM UTC-8, smn wrote:

> On Wednesday, February 20, 2013 8:13:24 PM UTC-8, William Elliot wrote:

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> > A problem from

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> > http://at.yorku.ca/cgi-bin/bbqa?forum=ask_an_analyst&task=list

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> > Suppose f(x)= e^(-1/x^2) for x not equal to 0, and f(0)=0.

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> > Without using l'hopital's rule, prove f is differentiable at 0 and

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> > that f'(0)=0.

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> for y>0 let g(y)=ye^(-y) ;its derivative has a 0 at y=1 ;is increasing for y<1 and decreasing for y>1 (eg the 2nd derivative of g is positive for all y>0.

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> Thus for y>1 , g(y)<g(1) so g(y)/y^1/2 = y^1/2 e^(-y)--> 0 as y---> +oo

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> Put 1/x^2 for y and let x-->0 (x>0) to get your result (the case x<0 follows since what we proved means that |f(x)/x|--> 0 as x-->0 (or |x|-->0). smn