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finding points of discontinuity of the function $f(x) = e^{x + (1/x)} $ and state their types.

My answer: this function has an essential discontinuity at 0, am I correct?

Ѕᴀᴀᴅ
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  • The function is undefined at $0$ – ℋolo Oct 05 '18 at 00:01
  • yes this is what I said ...... are there any other points at which the function is undefined? @Holo –  Oct 05 '18 at 00:02
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    you said it is "discontinuous" at 0, there is a difference between discontinuous and undefined(and no, there are no other points which this function undefined) – ℋolo Oct 05 '18 at 00:03
  • correct but discontinuities are found at points where the function is undefined correct? @holo ...... so my answer is correct? –  Oct 05 '18 at 00:06
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    no, look here to see discussion about this. But generally, a function is neither discontinuous nor continuous at points it does not defined. – ℋolo Oct 05 '18 at 00:10
  • @Happy The "functions" tag is used for questions about properties of functions/mappings in a set-theoretical frame such as injectivity and surjectivity. Please do not abuse this tag. – Ѕᴀᴀᴅ Oct 05 '18 at 00:26
  • okay I am sorry I will delete it @AlexFrancisco –  Oct 05 '18 at 00:29
  • Do you mean “singularity” instead of “discontinuity”? – Michael Hoppe Oct 05 '18 at 10:30
  • @MichaelHoppe No I mean discontinuity. –  Oct 06 '18 at 05:41
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    Your answer is correct. There are no other discontunuity points. – Harshit Joshi Oct 06 '18 at 07:03

1 Answers1

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A key data about $f$ is missong: its domain and codomain.

Anyway. It is simpler to deal with $e^{1/x}$ and, once you have understood it, may be multiply it by $e^x$ and then try to apply L'Hospital rule or something.

For complex numbers $z\in\mathbb{C}\setminus\{0\}$, the imaginary poles of the function $$f(z)=e^{1/z}$$ around $z=0$ are easily figured out: For any integer number $k$, $f$ has a pole at $$z_k=\frac{1}{2\pi ik}.$$ Now notice that, for any positive radius $r>0$, the domain of $f$ restricted to the open ball of radius $r$ centered at $z_\infty=0$ has an infinite number of poles. Thus $f$ has an essential singularity at $0$.

Now I ask you to complete the reasoning and comment: If $e^{1/z}$ has an essential singularity at $z=0$, then which type of singularity does $e^{z+1/z}$ has?

Dr Potato
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