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Let $g(x) = e^{-1/x^2}$ for $x$ not equal to zero, and $g(0) = 0$.

a) Please Show that $g^{(n)}(0) = 0$, for all $n = 0,1,2,3,4, \ldots$

Can someone please elaborate on the comments below for this one?

b) Please Show that the Taylor Series for $g$ about 0 agrees with $g$ only at $x = 0$.

I think this would be easy once I have part a, all I have to do is plug in n = 0?

Can someone please show how to do this?

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In a), do you mean "..., 8, 16, ...", or did you forget the 3? A more usual notation for the $n$-th derivative of $g$ is $g^{(n)}$ (with parentheses around the order). About the question itself: Once you have a), then b) is almost immediate, since the value of the Taylor series everyhwere is immediate from a) and it's a basic property of the exponential function that it can't take that value. For a), have you tried induction? – joriki Apr 8 '11 at 8:11
@joriki, The missing parenthesis is my fault. They were present in the original and I forgot to put them back in changing to LateX. – knucklebumpler Apr 8 '11 at 8:15
@joriki, I forgot the 3, sorry. For a, I have tried induction, but I messed up on the setup. – mary Apr 8 '11 at 8:52
@ Pete, Also, before you say "which is a very standard sort of limit" below, I don't know how to simplify it, the professor skipped l'hopitals rules and such – mary Apr 8 '11 at 8:54
@user8917: I added the $3$; usually, if someone points out an error in your question (or answer), it's good style to correct it so that the question can be understood without reading through all the comments. – joriki Apr 8 '11 at 9:12

A good way to do this is to prove something stronger. Let $P$ be any polynomial. Then show that the function $h$ defined by $h(x) = P(1/x)e^{-1/x^2}$ for $x \ne 0$ and $h(x)=0$ has your property $h^{(n)}(0) = 0$ for all $n$. Do it by showing: (1) $h$ is continuous; and (2) the derivative of such a function $h$ is another function of the same kind.

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All you have to do is prove that $\dfrac{\rm{d}g(x)}{\rm{d}x} = (\text{something})\,g(x)$ so that the limit of any derivative when $x\rightarrow 0$ is zero, since g(0)=0.

I see that

$$\dfrac{\rm{d}g(x)}{\rm{d}x} = \dfrac{2}{x^3}\,g(x) $$

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ok, that doesn't really answer part a fully. For part b, this is not sufficient and I am unsure what to do, can you please show it? – mary Apr 10 '11 at 4:05
This does no good at all if the (something) blows up as $x$ approaches 0 - which in fact the first derivative does, as per the formula you give. You need the fact that $g(x)$ approaches 0 'exponentially' as $x$ approaches 0, and this essentially boils down to GEdgar's method. – Steven Stadnicki Apr 10 '11 at 4:06

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