# is $\exp \left( - 1 \over x^2 \right)$ differentiable at $x=0$

is $\exp \left( - 1 \over x^2 \right )$ differentiate at $x=0$. Wolframlapha says it is. But is it continuous since we have $1 \over x^2$ and $x=0$? Can we really do this $f(0) = \exp \left( - {1 \over 0} \right)$? I hope I'm making any sense.

EDIT:: Does differentiability necessitate continuity? Is above function continuous at $x=0$.
Also a function $f(x) = {|x| \over x}$ also seems to be differentiable at $x=0$ as mentioned below in comment.

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It is a common misconception that continuity has something to do with evaluating at a point. It is nothing to do with this, it is about how the function behaves "approaching" a point. –  fretty Sep 20 '12 at 8:18
@fretty doesn't differentiability necessiate contuinity? –  Monkey D. Luffy Sep 20 '12 at 8:19
Yes but you show in your post that you do not understand continuity... –  fretty Sep 20 '12 at 8:29
The fact that you cannot evaluate $f$ at $0$ does not imply discontinuity. –  fretty Sep 20 '12 at 8:31
Depends on how strict your definition of differentiable is. You can just add into the definition $f(0)=0$ and everything will be fine by every definition. –  Nick Alger Sep 20 '12 at 8:31
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The function $$x \mapsto e^{-\frac{1}{x^2}}$$ can be extended in a continuous way to $x=0$. This extended function turns out to be differentiable (and even $C^\infty$). Strictly speaking, the original function is undefined at $x=0$, so that it is perfectly meaningless to ask whether it is differentiable.
@MonkeyD.Luffy In this case, you simply put $f(0)=0$. –  Siminore Sep 20 '12 at 9:17
Actually, you could define a derivative which also is well defined at isolated undefined points, provided that the function can be extended to an ordinarily differentiable function there: Just use $\lim_{h\to 0}\frac{f(x+h)-f(x-h)}{2h}$ instead of the standard definition. Indeed, with that definition, you could even derive non-continuous functions, as long as they only differ from continuous ones in isolated points (this is similar to the Lebesgue integral which also doesn't care about localized changes, although there you've got even more freedom). –  celtschk Sep 20 '12 at 9:38