
Re: Differentiability
Posted:
Feb 24, 2013 1:15 PM


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

