# Infinitely differentiable function

by IniquiTrance
Tags: differentiable, function, infinitely
 P: 183 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.
P: 4,572
 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).
 P: 183 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.
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,294 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.
 P: 81 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
Emeritus