Thread: Derivative of exp(-x^-2) at 0

1. Derivative of exp(-x^-2) at 0

The problem 38, from Calculus with Analytic Geometry by Purcell and Varberg, 5th edition, chapter 11, section 8 reads, in part:

Let

$$f(x) = \begin{cases}e^{-1/x^2}\quad x ≠ 0\\0\quad\quad\quad x = 0\end{cases}$$

(a) Show that $$f'(0) = 0$$ by using the definition of the derivative.

So I wrote:

$$f'(0) = \lim_{h \to 0}\frac{e^{-1/h^2} - 0}{h}$$

This gives a 0/0 indeterminate form, which usually calls for l'Hôpital's Rule, but that would involve determining the derivative of the function whose derivative I meant to determine by the above method.

Applying l'Hôpital's Rule gives $$\lim_{h \to 0}\frac{2e^{-1/h^2}}{h^3}$$ which makes things worse, as further application of l'Hôpital's Rule on the above will confirm.

It does work with the following adjustment:

$$f'(0) = \lim_{h \to 0}\frac{e^{-1/h^2}}{h} = \lim_{h \to 0}\frac{\frac{1}{h}}{e^{1/h^2}} = \lim_{h \to 0}\frac{-\frac{1}{h^2}}{-\frac{2}{h^3}e^{1/h^2}} = \lim_{h \to 0}\frac{h}{2}e^{-1/h^2} = 0$$

But I still don't feel like I solved the problem.

2. Re: Derivative of exp(-x^-2) at 0

\begin{align}f'(0) &= \lim_{h \to 0}\frac{e^{-1/h^2} - 0}{h} \\ &= \lim_{h \to 0}\frac1h e^{-1/h^2} \\ &= \lim_{k \to \pm\infty} \frac{k} {e^{k^2}} \\ &= \lim_{k \to \pm\infty} \frac1k \cdot \frac{k^2} {e^{k^2}} \\ &= \cancelto{0}{\left( \lim_{k \to \pm\infty} \frac1k\right)} \cdot \cancelto{0}{\left( \lim_{j \to +\infty} \frac{j} {e^{j}} \right)} \\ &= 0 \end{align}
The final limit (in $j$) is a standard result. You can get away with less detail than there is here.

In general, if you are proving a derivative, I would avoid the use of derivatives.