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

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William Elliot

Posts: 1,480
Registered: 1/8/12
Re: Differentiability
Posted: Feb 23, 2013 9:53 PM
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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



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