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Let $f : \mathbb{R} \to \mathbb{R}$ be the function $f(t) = \exp\left(- \frac{1}{t}\right) {\chi}_{(0 , \infty)}(t)$ for all $t \in \mathbb{R}$. I have to show that, for all $k \in \mathbb{N}$, exists a polynomial $P_k$ such that $$ f^{(k)}(t) = P_k\left(\frac{1}{t}\right) \exp\left(- \frac{1}{t}\right) \tag{1} $$ for all $t \in (0 , \infty)$, and also I have to see that $f^{(k)}(0)$ exists and it's equal to $0$ for all $k \in \mathbb{N}$. I have already shown $(1)$ by induction principle, but I haven't shown that $f^{(k)}(0) = 0$ but I have tried induction too: it's clair that $f(0) = 0$ by definition, and also $f'(0) = 0$, because, by L'Hôpital and using the continuity of $f$, $$ f'(0) = \lim_{t \to 0^+} \frac{f(t) - f(0)}{t - 0} = \lim_{t \to 0^+} \frac{\exp\left(- \frac{1}{t}\right)}{t} = \lim_{t \to 0^+} \frac{\frac{1}{t}}{\frac{1}{\exp\left(- \frac{1}{t}\right)}} = $$ $$ = \lim_{t \to 0^+} \frac{- \frac{1}{t^2}}{- \frac{1}{t^2} {\left(\exp\left(- \frac{1}{t}\right)\right)}^{- 2}} = \lim_{t \to 0^+} {\left(\exp\left(- \frac{1}{t}\right)\right)}^2 = 0^2 = 0\mbox{.} $$ Now, fix $k \in \mathbb{N}$ and let's suppose that $f^{(k)}(0) = 0$ to see that $f^{(k + 1)}(0) = 0$ too, so we have that $$ f^{(k + 1)}(0) = {(f^{(k)})}'(0) = \lim_{t \to 0^+} \frac{f^{(k)}(t) - f^{(k)}(0)}{t - 0} = \lim_{t \to 0^+} \frac{P_k\left(\frac{1}{t}\right) \exp\left(- \frac{1}{t}\right)}{t} = \lim_{t \to 0^+} \frac{\frac{1}{t}}{\frac{1}{P_k(t) \exp\left(- \frac{1}{t}\right)}} = $$ $$ = \lim_{t \to 0^+} \frac{- \frac{1}{t^2}}{- \frac{- \frac{1}{t^2} {P_k}'\left(\frac{1}{t}\right) \exp\left(- \frac{1}{t}\right) + \frac{1}{t^2} P_k\left(\frac{1}{t}\right) \exp\left(- \frac{1}{t}\right)}{{\left(P_k\left(\frac{1}{t}\right) \exp\left(- \frac{1}{t}\right)\right)}^2}} = \lim_{t \to 0^+} \frac{{\left(P_k\left(\frac{1}{t}\right) \exp\left(- \frac{1}{t}\right)\right)}^2}{\left(P_k\left(\frac{1}{t}\right) - {P_k}'\left(\frac{1}{t}\right)\right) \exp\left(- \frac{1}{t}\right)}\mbox{.} $$ My problem now is how to show that $$ \lim_{t \to 0^+} \frac{P_k^2\left(\frac{1}{t}\right) \exp\left(- \frac{1}{t}\right)}{P_k\left(\frac{1}{t}\right) - {P_k}'\left(\frac{1}{t}\right)} = 0\mbox{.} \tag{2} $$ Later, I can define $\psi : {\mathbb{R}}^n \to \mathbb{R}$ by $\psi(x) = \exp\left(- \frac{1}{1 - {\|x\|}^2}\right) {\chi}_{B(0 , 1)}$, being $B(0 , 1)$ the unit open ball in ${\mathbb{R}}^n$ and I would show that $\psi \in {\mathcal{C}}_c^{\infty}({\mathbb{R}}^n)$ but I have to show first $(2)$. Thank you very much.

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    Is ${\chi}_{(0 , \infty)}$ but the limits are $t \longrightarrow 0^+$ instead of $t \longrightarrow 0^-$. Thanks – 2017-01-27

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Notice that $$ \frac{f^{(k)}(t)}t = Q_k\left(\frac{1}{t}\right) \exp\left(- \frac{1}{t}\right) $$ for some polynomial $Q_k$. Hence, $$ \lim_{t\to0^+}\frac{f^{(k)}(t)}t=0 $$ (since the exponential "wins" to the polynomial).

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    I can not give a more rigorous reasoning since I do not know explicitly the expression of the polynomial. Do you think that? – 2017-01-27
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    I think that this is the cleanest way. But you need to take $\chi_{(-\infty,0)}$ instead of $\chi_{(0,+\infty)}$, think about it. – 2017-01-27
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    Why do I need ${\chi}_{(- \infty , 0)}$? – 2017-01-27
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    You don't need to take limits where the function is constant equal to zero. So, either $(0,+\infty)$ and $t\to0^-$ or $(-\infty,0)$ and $t\to0^+$. – 2017-01-27
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    Yes, before I said that $t \longrightarrow 0^+$ instead of $t \longrightarrow 0^-$. Thank you by your help. – 2017-01-27