In the current assignment for real analysis, we consider the following function

\[

f(x) =

\begin{cases}

e^{-1/x^2} & x \neq 0\

0 & x = 0.

\end{cases}

\]

We are asked to show that \(f^{(n)}(0) = 0\) for all \(n\), and that the Taylor series for \(f\) at \(0\) converges to \(f(x)\) only for \(x \neq 0\).

To compute the derivatives \(f^{(n)}(0)\), it is easiest to use induction on \(n\). For \(n = 1\),

\[

f'(0) = \lim_{x \to 0} \frac{f(x) – f(0)}{x – 0} = \lim_{x \to 0} \frac 1 x e^{-1 / x^2}.

\]

To compute the limit, we first use the change of variables \(y = 1 / x\) so that as \(y \to \pm \infty\) we have \(x \to 0\). Then

\[

\lim_{x \to 0} \frac 1 x e^{-1 / x^2} = \lim_{y \to \pm \infty} \frac{y}{e^{y^2}}.

\]

Using L’Hopital’s rule, it is easy to verify that the latter limit is zero. In fact, a similar argument shows that for any \(k\)

\[

\lim_{x \to 0} \frac{1}{x^k} e^{-1 / x^2} = 0.

\]

We will need this fact later on. At any rate, we’ve shown that \(f'(0) = 0\).

For the inductive step, assume that \(f^{(n)}(0) = 0\). Then

\[

f^{(n + 1)}(0) = \lim_{x \to 0} \frac{f^{(n)}(x) – f^{(n)}(0)}{x – 0} = \lim_{x \to 0} \frac{1}{x} f^{(n)}(x).

\]

After computing the first few derivatives, \(f'(x),\ f”(x), \ldots\) for \(x \neq 0\), you should be convinced that \(f'(x)\) is of the form

\[

f^{(n)}(x) = \left (\frac{a_k}{x^k} + \frac{a_{k-1}}{x^{k-1}} + \cdots + a_0 \right ) e^{-1 / x^2}.

\]

Using the same trick as before, we can compute

\[

\lim_{x \to 0} \frac{a_i}{x^{i+1}} e^{-1 / x^2} = 0,

\]

hence

\[

f^{(n + 1)}(0) = \lim_{x \to 0} \frac{1}{x} f^{(n)}(x) = 0.

\]

Since all of the derivatives of \(f\) are \(0\) at \(x = 0\), the Taylor series there is

\[

\sum_{k = 0}^\infty f^{(k)}(0) x^k = 0.

\]

However, \(f(x) \neq 0\) for \(x \neq 0\), so the Taylor series only agrees with \(f\) at \(x = 0\).