# Discontinuity points. Who was born first? The egg or the chicken?

Hi I started doing the "Calculus in a Nutshell" because even if I passed the exam at university I still have a lot of doubts about my understanding. So when I answered this question I had a new doubt. As you can see the Function since has x at denominator has to have a point of discontinuity for x=0. But since this function is equivalent to a straight line if I only see the straight line function I should have no discontinuity points. So my question is... If someone shows me a function like the one in the answer and ask me if there are discontinuity points and I answer that there are none I am wrong because I can lead back the function to the one with a denominator? Or it's just a matter of who was born first? The egg or the chicken?

I apologize for my bad English.

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1 month ago

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Is it possible to give down the function you had difficulties with? (You haven’t given the function in the note)

- 1 month ago

I posted a screenshot in the post but I'll write it down anyway the function is $f(x)=\frac{(x+a)^2-a^2}{x}$ can be rewritten as $f(x)=x+2a$ with $x\neq0$

- 1 month ago

- 1 month ago

If someone were to show you the function $f(x) = x + 2a$, you would certainly be correct in saying that there are no discontinuity points. However, if someone were to show you the function $f(x) = x + 2a, ~\boldsymbol{x \neq 0}$, then you would not be able to give a correct answer. Whoever was showing you this function would be guilty of not giving you enough information. In other words, they would be asking you a question about all points on $f(x)$, but not telling you what happens when $x = 0$. So it would be impossible to give an answer. You could perhaps guess that there was a discontinuity point, since they told you that $f(x)$ was only correct when $x \neq 0$, which is "suspicious". But the only way to indicate that there is a discontinuity at $x = 0$ is to give the full function, $f(x) = \frac{(a+x)^2 - a^2}{x}$ . Hope that helps!

- 1 month ago

Yup it helped. Thank you very very much!

- 1 month ago

No problem!

- 1 month ago

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