Show that $$f(x)= \begin{cases} exp(-\frac{1}{x^2}), & \text{if $x\gt 0$} \\[2ex] 0, & \text{if $x\le 0$ } \end{cases}$$

is infinitely differentiable.

Clearly $f^{(n)}(x)=0$ for all $x\lt 0$ and $f_{-}^{(n)}(0)=0$. Also since the derivatives of $exp(-1/x^2)$ produce $exp(-1/x^2)$ and a polynomial in $1/x$, using the rules for differentiation we can evaluate $f^{(n)}$ if $x\gt 0$ with any $n$. It remains to show that $f_{+}^{(n)}(0)=0$.

Now this final part is where I'm struggling. How can I show this part? I would greatly appreciate any help.

  • $\begingroup$ I find it useful to prove a lemma that any rational polynomial $\frac{p(x)}{q(x)}$ times $f(x)$ approaches zero as $x$ approaches zero. $\endgroup$ – Michael Burr Aug 17 '15 at 18:09
  • $\begingroup$ I see that is equivalent to showing that for any $n$, $f(x) 1/x^n$ approaches zero as $x$ approaches zero. But how can I show this? I've done it for $n=1$ but can't show it for the general case. $\endgroup$ – nomadicmathematician Aug 17 '15 at 18:21

Hint: Use induction and l'Hopital. Consider $\lim_{x\rightarrow 0^+}\frac{e^{-1/x^2}}{x^2}$. Let $y=1/x$, so this becomes $\lim_{y\rightarrow\infty}\frac{e^{-y^2}}{y^{-2}}=\lim_{y\rightarrow\infty}\frac{y^2}{e^{y^2}}$. After one step of l'Hopital, you have your answer. This approach can be generalized for any $n\geq 0$.

  • $\begingroup$ I see how it works!! But isn't there another way without using l'Hopital? This problem appears in a section before l'Hopital. $\endgroup$ – nomadicmathematician Aug 17 '15 at 18:40
  • $\begingroup$ You could also write $\frac{y^2}{e^{y^2}}=e^{2\ln y-y^2}$ and $2\ln y-y^2$ approaches $-\infty$ as $y$ increases. $\endgroup$ – Michael Burr Aug 17 '15 at 18:47

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