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I need help with this task, if anyone had a similar problem it would help me !

The task is: Determine the type of interruption at the point x = 0 for the function

$$f(x)=2^{-\frac{1}{x^{2}}}$$

I did:

$$L=\lim_{x\to 0^{-}} 2^{-\frac{1}{x^{2}}} = 0 $$ $$R=\lim_{x\to 0^{+}} 2^{-\frac{1}{x^{2}}} = 0 $$ $$L=R=x=0$$

And as I concluded the function is continuous at x = 0, but in the solution it says that the break is of the first kind. So I don’t understand why a breakup is the first kind?

Thanks in advance !

LogicNotFound
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  • I have never seen the terminology "break of the first kind". Can you provide the definition ? –  Mar 10 '21 at 09:05

3 Answers3

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The real function $x\mapsto 2^{-x^{-2}}$ has a removable discontinuity at $0$, there is no doubt about it.

  • Continuity condition: $$\lim_{x\to a^{-}} f(x)= \lim_{x\to a^{+}} f(x)=f(a) ?$$ I don't see how that's not the case here? – LogicNotFound Mar 10 '21 at 08:50
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    @SrdjanPesevic What's $f(0)$ here? Certainly not $2^{-0^{-2}}$. –  Mar 10 '21 at 09:09
  • $$f(0)$$ here is : $$f(0)=2^{-1\cdot 0^{-2}}=1 ?$$ And that's why they says that the break is of the first kind. Thank you so much ! – LogicNotFound Mar 10 '21 at 09:16
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    @SrdjanPesevic There is no such number as $2^{-1\cdot 0^{-2}}$ and surely there isn't any reason to claim it should be $1$. Anyways, if you want to stand by the claim that $f(0)$ is defined to be $1$, then $f$ is discontinuous at $0$, as you've shown in your post (I don't know eactly how you classify that; I would still call it a removable discontinuity). –  Mar 10 '21 at 09:20
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You have a removable discontinuity at $x = 0$.

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After OP's calculation of L and R which are equal to zero.

A prescription for $f(0)$ is required. if $f(0)=K=0$, the function is continuous at $x=0$. If $K\ne 0$, the function $F(x)$ is discontinuous at $x=0$. So there is a removable discontinuity at $x=0$ which can be avoided by choosing $K=0$.

Z Ahmed
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