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Topic: Differentiability
Replies: 8   Last Post: Mar 1, 2013 6:22 PM

 Messages: [ Previous | Next ]
 Stuart M Newberger Posts: 479 Registered: 1/25/05
Re: Differentiability
Posted: Mar 1, 2013 6:22 PM

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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> > 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

Date Subject Author
2/20/13 William Elliot
2/21/13 Robin Chapman
2/21/13 David C. Ullrich
2/22/13 William Elliot
2/23/13 David C. Ullrich
2/23/13 William Elliot
2/24/13 David C. Ullrich
2/28/13 Stuart M Newberger
3/1/13 Stuart M Newberger