On Fri, 22 Feb 2013 00:34:12 -0800, William Elliot <firstname.lastname@example.org> wrote:
>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.
What I wrote is so obviously false that you might consider the possibility that it was a typo or something, and think about what the correct version is. Of course I meant
(**) 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 >> 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.
Which is no help with your question. What does (**) say about e^(-x) for x > 0?