On Wednesday, February 20, 2013 8:13:24 PM UTC-8, William Elliot wrote: > A problem from > > http://at.yorku.ca/cgi-bin/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 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