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\textbf{Extra credit homework 13, due December 4} \\
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I've included solutions for each part.

Consider the function $f(x)$, defined to be
\[
f(x) = \begin{cases}
e^{-1/x^{2}} & \text{when $x \neq 0$,} \\
0 & \text{when $x=0$}.
\end{cases}
\]

(a) Show that $f'(0) = 0$.

\textbf{Solution:} To compute $f'(0)$, you need to use the definition
of derivative:
\begin{align*}
f'(0) &= \lim_{h \rightarrow 0} \frac{f(h) - f(0)}{h} \\
      &= \lim_{h \rightarrow 0} \frac{e^{-1/h^{2}} - 0}{h} \\
      &= \lim_{h \rightarrow 0} \frac{e^{-1/h^{2}}}{h} \\
      &= \lim_{h \rightarrow 0} \frac{1/h}{e^{1/h^{2}}}.
\end{align*}
(Technical point: in this limit, we can assume that $h$ is never 0,
which is why I can write $f(h) = e^{-1/h^{2}}$.)  Now use
L'H\^opital's rule:
\begin{align*}
      &= \lim_{h \rightarrow 0} \frac{-1/h^{2}}{-2/h^{3} e^{1/h^{2}}} \\
      &= \lim_{h \rightarrow 0} \frac{h}{2 e^{1/h^{2}}} \\
      &= 0.
\end{align*}
(This last limit is zero because the numerator approaches zero, while
the denominator approaches $\infty$.)

(b) Show that $f''(0) = 0$.

\textbf{Solution:} This is similar.  First, we need to compute $f'(x)$
when $x \neq 0$: we just apply the chain rule to $e^{-1/x^{2}} =
e^{-x^{-2}}$: $f'(x) = \frac{2}{x^{3}} e^{-1/x^{2}}$.  So:
\begin{align*}
f''(0) &= \lim_{h \rightarrow 0} \frac{f'(h) - f'(0)}{h} \\
       &= \lim_{h \rightarrow 0} \frac{\frac{2}{h^{3}} e^{-1/h^{2}}- 0}{h} \\
       &= \lim_{h \rightarrow 0} \frac{2 e^{-1/h^{2}}}{h^{4}} \\
       &= \lim_{h \rightarrow 0} \frac{2 h^{-4}}{e^{1/h^{2}}} \\
       &= \lim_{h \rightarrow 0} \frac{-8 h^{-5}}{-2h^{-3}
e^{1/h^{2}}}   \ \ \text{(L'H\^opital)}\\
       &= \lim_{h \rightarrow 0} \frac{4 h^{-2}}{e^{1/h^{2}}} \\
       &= \lim_{h \rightarrow 0} \frac{-8 h^{-3}}{-2h^{-3}
e^{1/h^{2}}} \ \ \text{(L'H\^opital)}\\
       &= \lim_{h \rightarrow 0} \frac{4}{e^{1/h^{2}}} \\
       &= 0.
\end{align*}

(c) Show that $f^{(n)}(0) = 0$ for as many values of $n$ as you can.
Ideally, explain why this is zero for all $n$.

\textbf{Solution:} I'm not going to give the details.  One approach
would be to show that 
\[
\lim_{h \rightarrow 0} \frac{e^{-1/h^{2}}}{h^{k}} = 0
\]
for any positive integer $k$, and then show that $f^{(n)}(0)$ is
computed as a sum of such limits.
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