On Thu, 21 Feb 2013, David C. Ullrich wrote: > > 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. > > You really _should_ try proving this using l'Hopital and see what > happens... > f'(0) = lim(x->0) (exp -1/x^2)/x = lim(x->0) 2(exp -1/x^2)/x^3 = lim(x->0) 4(exp -1/x^2)/x^3 * 1/2x^2
> Hint: One way or another, show that > > (*) e^x < x (x > 0). > Can't be done. e^x = 1 + x + x^2 / 2 + ... > x for x > 0.
> For example, using the power series, or using the fact that > e^x - 1 = int_0^x e^t dt > or whatever. > Oh?
> Now what does (*) imply about e^(-x) for x > 0?
Since Dexp is positive, exp is increasing and exp -x decreasing. So for x > 0 exp -x < exp 0 = 1.