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

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David C. Ullrich

Posts: 21,553
Registered: 12/6/04
Re: Differentiability
Posted: Feb 24, 2013 1:15 PM
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On Sat, 23 Feb 2013 18:53:34 -0800, William Elliot <marsh@panix.com>
wrote:

>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] 1/x < exp -x

At first I assumed this was just a typo. But no, [1] is
exactly what you use below, so it must be what you
meant.

Inequality [1] is obvious nonsense. Consider small x > 0
and you see [1] implies ifinity < 0.

What actually follows from (**) is

[2] 1/x > exp(-x) (x > 0).

Because 0 < a < b implies 1/a > 1/b. Like 2 < 3
so 1/2 > 1/3.

>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


Since [1] is wrong and you use [1] here I assumed at
first there must be an error. But no, this chain of
inequalities is correct (or rather would be correct if [1]
were correct).

The actual proof, using [2]: For any x <> 0, positive
or negative, [2] shows that

exp(-1/x^2) < x^2.

So

|exp(-1/x^2)/x| < x^2/|x| = |x|,

so the limit is 0.

Now for a minute I was puzzled how both [2] and
its opposite [1] could lead to a proof that f'(0) = 0.
The answer: You applied [1] for small x, while the
correct proof applies [2] for large x.

>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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