
Re: Differentiability
Posted:
Feb 23, 2013 9:53 PM


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
f'(0+) = lim(x>0+) (exp 1/x^2)/x = 0 Proof 0 <= lim(x>0+) (exp 1/x^2)/x = lim(x>0+) x(exp 1/x^2)/x^2 <= lim(x>0+) x(exp 1/x^2)^2 = 0
f'(0) = lim(x>0) (exp 1/x^2)/x = lim(x>0+) (exp 1/x^2)/x = 0 f'(0) = f'(0+) = f'(0) = 0

