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

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Stuart M Newberger

Posts: 479
Registered: 1/25/05
Re: Differentiability
Posted: Feb 28, 2013 11:33 PM
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On Wednesday, February 20, 2013 8:13:24 PM UTC-8, William Elliot wrote:
> A problem from
> Suppose f(x)= e^(-1/x^2) for x not equal to 0, and f(0)=0.
> Without using l'hopital's rule, prove f is differentiable at 0 and
> that f'(0)=0.

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 3nd derivative of g is positive for all y>0.
Thus for y>1 , g(y)<g(1) so g(y)/y^1/2 = y^1/2 e^(-y)--> 0 as y---> +oo
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

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