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Topic: Differentiability
Replies: 8   Last Post: Mar 1, 2013 6:22 PM

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William Elliot

Posts: 1,610
Registered: 1/8/12
Re: Differentiability
Posted: Feb 22, 2013 3:34 AM
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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.



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