what is the value of the function $\exp(-\frac{1}{x^2})$ at the origin?

Question

I have found in Boas' book that the function $\exp(-\frac{1}{x^2}$) and its derivatives are zero at the origin. But when I evaluated the first derivative of the function, I found something like this: $$\frac{d}{dx}\left(\exp(-\frac{1}{x^{2}})\right)=\frac{2}{x^{3}\exp\left(\frac{1}{x^{2}}\right)}$$ Now if my exercise is correct, I think that the value of this should be an indeterminate rather than zero. I need some help in understanding this...

Answer 1

You have to find these limits in indeterminate form using L'Hopital's Rule. This rule can be used repeatedly to show that $\frac {e^{x}} {x^{n}} \to \infty $ as $ x\to \infty $ for any positive integer $n$. This should help you to to prove the stated property of $e^{-1/{x^{2}}}$. Hint: all derivatives of $e^{-1/{x^{2}}}$ are of the type $p(\frac 1 x) e^{-1/{x^{2}}}$ where $p$ is a polynomial, as seen by an induction argument.

Answer 2

The function you have in mind is $$ f(x)=\begin{cases} e^{-1/x^2} & x\ne 0 \\[6px] 0 & x=0 \end{cases} $$ Since $$ \lim_{x\to0}e^{-1/x^2}=0 $$ the function is continuous at $0$. The derivative can be computed by “first principles”: $$ \lim_{x\to0}\frac{f(x)-f(0)}{x-0}= \lim_{x\to0}\frac{e^{-1/x^2}}{x}= \lim_{x\to0}x\frac{e^{-1/x^2}}{x^2} $$ Now $$ \lim_{x\to0}\frac{e^{-1/x^2}}{x^2}= \lim_{t\to\infty}te^{-t}= \lim_{t\to\infty}\frac{t}{e^t}=0 $$ so also $$ \lim_{x\to0}x\frac{e^{-1/x^2}}{x^2}=0\cdot0=0 $$

Answer 3

By doing a small trick of defining auxiliary variable $t = 1/x$, you can split your problem into two ones: $t$ tends to $+\infty$ and $-\infty$. Moreover, if you realize the symmetry of your function, two new problems become one.

Answer 4

You should notice that the domain of $f(x)=e^{-\frac{1}{x^2}}$ is $x \neq 0$. In another word, it has no definition at the original point. Thus, it also has no any derivative there.

But since $$\lim_{x \to 0}e^{-\frac{1}{x^2}}=\lim_{y \to -\infty}e^y=0,$$thus, if we complement to define $f(0)=0$, then $f(x)$ is continuous everywhere.

Consider the derivative $f'(0)$. At this moment, you can't evaluate it directy by the derivation formula.(Why?) And you only can do that by the definition of the derivative as follows

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