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I only know that $g(x)=\lim_{x \downarrow 0}e^{-\frac{1}{x}}=0$ and that $g(x)$ is smooth and all the derivatives vanish at $x=0$. How can I conclude from this the result above?

Max
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New2Math
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2 Answers2

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Let $f$ be given by

$$f(x)=e^{-1/(b-x)}\tag1$$

for $x<b$

It is straightforward to show that, for $x<b$, the nth derivative, $f^{(n)}(x)$ of $f(x)$ can be expressed as

$$f^{(n)}(x)=g_n(x)e^{-1/(b-x)}\tag2$$

where in $(2)$, $g_n(x)$ is a polynomial of order $2n$ in powers of $1/(b-x)$. The sequence of functions, $g_n(x)$, satisfies the relationship

$$g_{n+1}(x)=g_n'(x)-\frac1{(b-x)^2}g_n(x)$$

with $g_0(x)=1$.

A closed-form for $g_n(x)$ can be found using the Faà_di_Bruno Formula. We do note need, however, a closed-form for $g_n(x)$ to proceed since clearly, we have $$\lim_{x\to b^-}g_n(x)e^{-1/(b-x)}=0$$

Mark Viola
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  • Thank you for your Reply I have understood the induction but I don't understand how you have evaluated the Limit. You have said that $g_n$ is a polynomial and its Arguments are $\frac{1}{b-x}$. How do you know that $\lim_{x\uparrow b}(\frac{1}{b-x})^{k}e^{-1/(b-x)}=0$, forall $k\in\mathbb{N}$? I know that $\lim_{y\downarrow 0}(\frac{1}{y})^{k}e^{-1/(y)}=0$, forall $k\in\mathbb{N}$ is there a possibility to derive the result from that somehow ? – New2Math Feb 14 '19 at 17:40
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    You're quite welcome. My pleasure. Note that for $b>x$ $$0<\frac{1}{(b-x)^k}e^{1/(b-x)}=\frac{1}{(b-x)^k} \frac1{e^{1/(b-x)}}\le \frac{1}{(b-x)^k}\left(\frac{1}{\frac{1}{(k+1)!(b-x)^{k+1}}}\right)=(k+1)!(b-x)$$ from exploiting the Taylor Series for $e^x$. Now, apply the squeeze theorem. – Mark Viola Feb 14 '19 at 17:45
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HINT

Change variables to $y = b-x$.

gt6989b
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