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I'm trying to prove that the function $f$ is of class $C^\infty$ as follows. Given any integer $n\geq 0$, define $f_n : \mathbb R \rightarrow \mathbb R$ by

$$f_n(x) = \frac{e^{-1/x}}{x^n}, \quad \text{for} \quad x > 0,$$

$$f_n(x) = 0, \quad \text{ for } \quad x<=0.$$

Question: How do show that $f_n$ is continuous at 0, differentiable at 0, and $f'_n(x) = f_{n+2}(x) -nf_{n+1}(x)$ for all $x$, and $f_n$ is of class $C^\infty$?

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First, I think you should work on the title of the question. Second, is this homework? If so, you should tag it as such. Also, I don't see why the standard things shouldn't work for this problem. What have you tried? Finally, you might elicit more/better responses if you go back and accept a few more of the answers to your previous questions... just a suggestion. :) –  William DeMeo Feb 22 '12 at 12:04
    
The title seems misleading. –  Neal Feb 22 '12 at 12:09
    
Not homework, just practice. I have tried differentiating the functions, but wasn't very organized –  James R. Feb 23 '12 at 5:53
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up vote 2 down vote accepted
  • We have $\lim_{x\to 0+}f_n(x)=\lim_{t\to+\infty}e^{-t}t^n=0=f_n(0)=\lim_{x\to 0^-}f_n(x)$, so $f_n$ is continuous at $0$.
  • The left derivative at $0$ is $0$ and a similar argument as what was done for continuity gives the differentiability (apply it to $n+1$ instead of $n$).
  • We have $f_n'(x)=f_{n+2}(x)-nf_{n+1}(x)$ if $x\leq 0$, and for $x>0$ we have $$f_n'(x)=\frac 1{x^2}e^{-1/x}\frac 1{x^n}-n\frac 1{x^{n+1}}e^{-1/x}=f_{n+2}(x)-nf_{n+1}(x).$$ Since $f_{n+2}$ and $f_{n+1}$ are differentiable, so is $f'_n$. So by induction if all $f_n$ are $C^k$ then all $f_n$ are $C^{k+1}$.
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