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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...

Jon
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StudentofPhysics
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  • In the denominator $\exp(1/x^2)$ grows so fast (when $x\to0$) that $x^3$ has no chance whatsoever to keep it bounded! The absolute value of that denominator $\to\infty$. – Jyrki Lahtonen Jul 11 '18 at 08:58
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    I think this question is close enough to be a duplicate of this more general version. I think your concerns are all covered there, but I may be wrong. – Jyrki Lahtonen Jul 11 '18 at 09:00
  • yes sir @JyrkiLahtonen. – StudentofPhysics Jul 11 '18 at 09:05
  • @Chris2006 I am rejecting your suggested edit. I think we should steer away from using displayed fractions in question titles. If others approve, then it will eventually be accepted. Writing this comment to explain my point of view. The "official" message accompanying my rejection is in my opinion unduly harsh, so I want to add a longer explanation. – Jyrki Lahtonen Jul 11 '18 at 09:49
  • @JyrkiLahtonen, Sir, I want to understand exactly what you meant when you said "x^3 has no chance whatsoever to keep it bounded". Sorry if it is too trivial a question. – StudentofPhysics Jul 11 '18 at 09:53
  • It can easily be proven that all derivatives are polynom of 1/x * the function itself. As the function goes to 0 faster than any polynom, all derivatives are null. – Pierre Jul 11 '18 at 10:24

4 Answers4

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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 $$ \lim_{x\to0}x\frac{e^{-1/x^2}}{x^2}=0\cdot0=0 $$

egreg
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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.

Kavi Rama Murthy
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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.

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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

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

WuKong
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