Showing that $f^{\left(n\right)}\left(0\right)\=0$ for $n\=1\,2\,\dots$, requires two things:

If $x\ne 0$, $f\prime \left(x\right)\=\frac{2{\ⅇ}^{-\frac{1}{x^2}}}{x^3}$, a result obtained by differentiating the rule for $f$ when $x\ne 0$.

To obtain $f\prime \left(0\right)$, apply the limit-definition of the derivative:

The actual value of the limit is obtained by applying L'Hôpital's rule to the indeterminate form $0\/0$, with the fraction written with $1\/h$ in the numerator.

$\frac{1\/h}{e^{1\/h^2}}\to \frac{-\frac{1}{h^2}}{-\frac{2}{h^3}{e}^{1\/h^2}}\=\frac{1}{2}h e^{-1\/h^2}\to 0$ 

 Thus, $f\prime \left(x\right)\=\left\{\begin{array}{cc}\frac{2e^{-1\/x^2}}{x^3}& x\ne 0\\ 0& x\=0\end{array}\right.$, a result needed to obtain $f″\left(0\right)$.

If $x\ne 0$, $f″\left(x\right)\=\left(\frac{4}{x^6}-\frac{6}{x^4}\right)e^{-1\/x^2}$, a result obtained by differentiating the rule for $f\prime \left(x\right)$ when $x\ne 0$.

To obtain $f″\left(0\right)$, evaluate $\underset{h\to 0}{lim}\frac{\frac{2e^{-1\/h^2}}{h^3}}{h}\=2\underset{h\to 0}{lim}\frac{e^{-1\/h^2}}{h^4}$. This limit also yields to L'Hôpital's rule if the numerator is taken as $1\/h^4$, and the rule is applied multiple times.

$\frac{h^{-4}}{e^{1\/h^2}}\to \frac{-4 h^{-5}}{-\frac{2}{h^3}{e}^{1\/h^2}}\to 2\frac{h^{-2}}{e^{1\/h^2}}\to 2 \frac{-2 h^{-3}}{-\frac{2}{h^3}{e}^{1\/h^2}}\=2 e^{-1\/h^2}\to 0$ 

Thus, $f″\left(x\right)\=\left\{\begin{array}{cc}\left(\frac{4}{x^6}-\frac{6}{x^4}\right)e^{-1\/x^2}& x\ne 0\\ 0& x\=0\end{array}\right.$, a result needed to obtain $f‴\left(0\right)$.

At this point, appropriate generalizations need to be made, both for the pattern of the derivatives and for the pattern of the limits. Each new derivative will require computing the limit of a fraction of the form $h^{-k}\/e^{1\/h^2}$, which will yield to multiple applications of L'Hôpital's rule.

The Taylor-series recipe applied to $f\left(x\right)$ given in this example is

$\sum _{n\=0}^{\infty }\frac{f^{\left(n\right)}\left(0\right)}{n\!}{x}^n$ $\=0\+0\+\cdots \=0$

a sum of terms all of which are zero, so that "series" is just zero. It does not represent $f\left(x\right)$ even though all the derivatives of $f\left(x\right)$ exist.

The code behind Maple's Limit Methods tutor proves most useful in applying L'Hôpital's rule multiple times. To make use of this code as painless as possible, it will be implemented in its command-form, thereby eliminating the tedium of launching the tutor repeatedly. Note also that the Rule command must be applied to the inert form of the limit operator. Conversion to inert is via the "2-D Math≻Convert To≻Inert Form" option in the Context Menu for the operator. The display of an inert operator is gray, rather than the default black.

These were the very calculations that were implemented in order that the Mathematical Solution would be correct. Here is an instance where Maple proves to be exceedingly helpful in establishing the pattern that has to be generalized.

An alternative to the use of the Rule command is the use of the Limit Methods tutor. Figure 8.5.1(a) shows the state of the tutor after L'Hôpital's rule is applied once to the limit $\underset{h\to 0}{lim}g\left(h\right)\/h^5$. Figure 8.5.1(b) shows how the rule is applied within the tutor.

The upper portion of Figure 8.5.1(b) shows the initial state of the rule. The numerator and denominator must be made to change places. Select, in turn, the numerator and denominator, and use the arrows to change their respective positions. The result is shown in the lower portion of Figure 8.5.1(b). When the "Apply" button is pressed, the tutor assumes the state shown in Figure 8.5.1(a).

L'Hôpital's rule is to be applied twice more in this way, resulting in a limit that yields to the Product rule.