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Re: Differentiability
Posted:
Feb 23, 2013 11:29 AM
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On Fri, 22 Feb 2013 00:34:12 -0800, William Elliot <marsh@panix.com>
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?
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