On Sat, 23 Feb 2013 18:53:34 -0800, William Elliot <firstname.lastname@example.org> wrote:
>On Sat, 23 Feb 2013, David C. Ullrich wrote: > >> >> > Suppose f(x)= e^(-1/x^2) for x not equal to 0, and f(0)=0. >> >> > prove f is differentiable at 0 and that f'(0)=0. > >> >> Hint: One way or another, show that > >> (**) e^x > x (x > 0). > >> >> For example, using the power series, or using the fact that >> >> e^x - 1 = int_0^x e^t dt > >> What does (**) say about e^(-x) for x > 0? >> >  1/x < exp -x
At first I assumed this was just a typo. But no,  is exactly what you use below, so it must be what you meant.
Inequality  is obvious nonsense. Consider small x > 0 and you see  implies ifinity < 0.
What actually follows from (**) is
 1/x > exp(-x) (x > 0).
Because 0 < a < b implies 1/a > 1/b. Like 2 < 3 so 1/2 > 1/3.