The bumpfunction $\psi$ has the properties that given two values $a,b\in\mathbb{R}$ such that $a<b$.
$\psi$ defines a smooth function with the properties
$$f(n) = \begin{cases} e^{-\frac{1}{x-a}}\cdot e^{-\frac{1}{b-x}} &\mbox{if } a<x<b \\ 0 & \mbox{otherwise } \end{cases} $$
and $\lim_{x\downarrow a}f^{(n)}(x)=0$ and $\lim_{x\uparrow b}f^{(n)}(x)=0$ ,$f^{(n)}$ describes the $n$-th derivative. $n\in\mathbb{N}_0$
I know that the $n-$th derivative for $e^{-\frac{1}{x-a}}=a(x)$ can be given by the Formula $q_n(-\frac{1}{x-a})a(x)$. The $n$-th derivative for $e^{-\frac{1}{b-x}} = b(x)$ can be given by the formula $p_n(-\frac{1}{b-x})$ Where $p_n$ and $q_n$ is a polynomial.
I know that the $n$-th derivative for $f(x)= (a\cdot b)(x)$ exists. Because since $a$ and $b$ are smooth the product is also smooth and the $n$-th derivative can given by the recursive fomula:
$f^{(1)}=(a\cdot b)^{(1)}=a'b+b'a$ and $f^{(n)}=(f^{(n-1)})'$
Now I want to prove that $\lim_{x\downarrow a}f^{(n)}(x)=0$ and $\lim_{x\uparrow b}f^{(n)}(x)=0$ , $n\in\mathbb{N}_0$
I have made the assumption that the $n$th derivative is of the form $$q_n(-\frac{1}{x-a})+p_n(-\frac{1}{b-x})((a\cdot b)(x))$$
But when I made the inductionstep I was runing into a Problem
$(a\cdot b)(x)((q_n(-\frac{1}{x-a})+p_n(-\frac{1}{b-x}))'+(((-\frac{1}{(x-a)^2})+(-\frac{1}{(b-x)^2}))(q_n(-\frac{1}{x-a})+p_n(-\frac{1}{b-x}))))$
The Problem is that if I multiply $-\frac{1}{(x-a)^2}$ with $p_n(-\frac{1}{b-x})$ then I get a polynomial with in a 'mixed' form. That's not what I want because this would negate the inductionstep.
Therefore I am asking you for help on the proof that:
$\lim_{x\downarrow a}f^{(n)}(x)=0$ and $\lim_{x\uparrow b}f^{(n)}(x)=0$ , $n\in\mathbb{N}_0$
