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Let $a<b$ and $f(x) = 0$ if $x$ is outside $(a,b)$ and $f(x) =e^{-\frac{1}{x-a}}e^{-\frac{1}{b-x}}$ is $a<x<b$. Show that $f$ is infinitely differentiable on $\mathbb R$

If $x\neq a,b$ then it is easy to see the result. I'm trying the cases $x=a$ and $x=b$. For $x=a$, we have $$ f_-'(a) = \lim_{h\to 0^-} \frac{f(a+h)-f(a)}{h} = \lim_{h\to 0^-} \frac{0-0}{h} = 0 $$ and $$ f_+'(a) = \lim_{h\to 0^+} \frac{e^{-\frac{1}{(a+h)-a}}e^{-\frac{1}{b-(a+h)}}-0}{h} = \frac{0}{0} $$ but using L'Hopital rule in this case we get stuck on the $0/0$. How can I proceed in this situation?

Thanks in advance.

Ain't No O
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1 Answers1

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\begin{align} f_+'(a) &= \lim_{h\to 0^+} \frac{e^{-\frac{1}{(a+h)-a}}e^{-\frac{1}{b-(a+h)}}-0}{h} = \\ &= e^{-\frac{1}{b-a}}\lim_{h\to 0^+} \frac{e^{-\frac{1}{h}}}{h} = \\ &= e^{-\frac{1}{b-a}}\lim_{k\to +\infty} ke^{-k} = \\ &= e^{-\frac{1}{b-a}}\lim_{k\to +\infty} \frac{k}{e^{k}} = \\ &= e^{-\frac{1}{b-a}}\lim_{k\to +\infty} \frac{1}{e^{k}} = 0 \end{align}

Vincenzo Tibullo
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