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Let f(x) = e^(-1/x) for x not equal to 0

      = 0 for all x=0

Find the value of the first, second, third, and fourth derivatives at x=0.

Intuitively, I feel like the value should be 0 for all derivatives, but I don't know how to prove it.

I also found the first, second, third, and fourth derivatives of e^(-1/x) just in case.

I found a cached copy of this similar question: http://webcache.googleusercontent.com/search?q=cache:https://math.stackexchange.com/questions/332142/looking-for-help-with-a-proof-that-n-th-derivative-of-e-frac-1x2-0-for

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Use the limit definition of the derivative

$$ \lim_{h\to 0} \frac{f(h) - f(0)}{h} = \lim_{h\to 0} \frac{f(h)}{h} $$

if $h < 0$, then $f(h) = 0$, and you're done. So it's sufficient to now consider

$$ \lim_{h\to 0+} \frac{f(h)}{h} = \lim_{h\to 0+} \frac{e^{-1/h}}{h}. $$ There are various ways you might evaluate this limit; one way is to let $y = 1/h$ and evaluate $$ \lim_{y \to \infty} \frac{e^{-y}}{1/y} = \lim_{y \to \infty} \frac{y}{e^{y}} $$ and then use L'Hopital's rule.

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