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Re: Differentiability
Posted:
Mar 1, 2013 6:22 PM


On Thursday, February 28, 2013 8:33:58 PM UTC8, smn wrote: > On Wednesday, February 20, 2013 8:13:24 PM UTC8, William Elliot wrote: > > > A problem from > > > > > > http://at.yorku.ca/cgibin/bbqa?forum=ask_an_analyst&task=list > > > > > > > > > > > > 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 2nd 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



