## Infinitely differentiable function

This might sound like a stupid question.

$f(x) = \begin{cases} &e^{-\frac{1}{x^2}} &\text{if } x\neq 0 \\ & 0 &\text{if } x = 0 \end{cases}$

Is the reason f is infinitely differentiable at 0 because we keep differentiating 0 as a constant, or because,

$\lim_{x\rightarrow 0} f(x) = \lim_{x\rightarrow 0} \frac{2}{x^3} e^{-\frac{1}{x^2}} = 0$

Thanks.
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 Quote by IniquiTrance This might sound like a stupid question. $f(x) = \begin{cases} &e^{-\frac{1}{x^2}} &\text{if } x\neq 0 \\ & 0 &\text{if } x = 0 \end{cases}$ Is the reason f is infinitely differentiable at 0 because we keep differentiating 0 as a constant, or because, $\lim_{x\rightarrow 0} f(x) = \lim_{x\rightarrow 0} \frac{2}{x^3} e^{-\frac{1}{x^2}} = 0$ Thanks.
Hey IniquiTrance and welcome to the forums.

If you want to show differentiability at a particular point from first principles, you need to show that the appropriate limit exists from both sides of the point (which is what you have done).

The only thing though that I would recommend is that you expand out the definition and show it step by step using the various limit theorems (since you are going to get infinities for this particular problem).
 Thanks for the reply chiro. I was just mainly wondering whether all order derivatives were 0, because we were taking the derivative of 0 as a constant which is uniformly 0, or instead because we are using the exponential and taking a limit. I'm assuming from your response that it is because of the latter case.

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## Infinitely differentiable function

One can show that the nth derivative is $e^{-1/x^2}$ over a polynomial in x. The numerator always goes to 0 faster than the denominator so the limit of the nth derivative, as x goes to 0, is 0. Much the same thing happens at x= 0. The "difference quotient, (f(h)- f(0))/h= f(h)/h, will be have $e^{-1/h^2}$ in the numerator and a polynomial in h in the denominator so the derivative at x= 0 is always 0 and every derivative is differentiable.
 You can apply L Hopital's rule setting $\epsilon$=[1][/x^2] so then $\epsilon$*($\epsilon$)^0.5*e^(-1*$\epsilon$) you can see this utilizing L'hopital's rule

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 Quote by IniquiTrance Thanks for the reply chiro. I was just mainly wondering whether all order derivatives were 0, because we were taking the derivative of 0 as a constant which is uniformly 0, or instead because we are using the exponential and taking a limit. I'm assuming from your response that it is because of the latter case.
It's the latter. Consider the function

f(x) = 0 if x=0, x if x is not zero. Is f'(0)=0 because we're taking the derivative of the constant function 0? Clearly not, right?
 You have to show that the function has a derivative at 0, and that it is 0. Also you need to show that the function is continuous(as that is required for it to be differentiable). This function is interesting because it obscures all the information about the e^(y) part of the function(and so doesn't have a taylor series expansion at 0).