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Prove or Disprove: Let $x_0\in\mathbb R$ and let $f$ be a function that is defined on a neighborhood of $x_0$. If $f$ is not continuous at $x_0$ and $f^3$ is continuous at $x_0$, then $f^2+f+1$ is not continuous at $x_0$.

I am struggling with this proof (or disproof), because in my mind if we have that $f^3$ is continuous at $x_0$, then $lim_{x\to x_0} f^3(x)=f^3(x_0)\implies lim_{x\to x_0}f(x)=\sqrt[\leftroot{-2}\uproot{2}3]{lim_{x\to x_0} f^3(x)}=\sqrt[\leftroot{-2}\uproot{2}3]{f^3(x_0)}=f(x_0)$ which just implies that $f$ is continuous at $x_0$ and then this is vacuously a proof.

Am I missing something?

yamyam263
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  • It could be possible if the values of $f$ are in $\mathbb{C}$ (or some other field without unique cube roots). – aschepler Dec 19 '21 at 14:32

2 Answers2

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Assume that it is, then $(f^{3}-1)/(f^{2}+f+1)=f-1$ is continuous at $x_{0}$, which in turn implies that of $f$, a contradiction.

user284331
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You are correct. The premises (are contradictory)[https://math.stackexchange.com/questions/3065701/can-a-cube-of-discontinuous-function-be-continuous/3065705] so the statement is trivially true.

Jsevillamol
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  • (Note however that generally you cannot take a power our of a limit like this ; otherwise you would be able to proof the same for square roots which is false) – Jsevillamol Dec 19 '21 at 14:27
  • Right but in this case, couldn't you say that if we assume that $f^3(x)$ approaches $f^3(x_0)$ as $x$ approaches $x_0$, then by substitution ($t:=f^3(x)$) can't we say that $f(x)$, which is $(f^3(x))^{1/3}$ or in other words $t^{1/3}$ approaches $f^3(x_0)^{1/3}$ as $t$ approaches $f^3(x_0)$? because that would be $f(x_0)$ which means we can use this transition in this case. But yes, it would reset the negative values for square roots. (And also it reset complex numbers) – yamyam263 Dec 19 '21 at 16:53