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At which points is the following function continuous?

$$\begin{eqnarray*} f(x) = \begin{cases} 5x, &\text{if }x \in\mathbb Q, \\ x^2-6, &\text{if }x \notin\mathbb Q. \end{cases} \end{eqnarray*}$$

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  • $\begingroup$ I've done nothing because I have no idea where to start from. $\endgroup$ – Gigili Jan 24 '12 at 11:00
  • $\begingroup$ $Q$ is the set of rational number? $\endgroup$ – Paul Jan 24 '12 at 11:01
  • $\begingroup$ When is $5x = x^2-6$? $\endgroup$ – lhf Jan 24 '12 at 11:01
  • $\begingroup$ Where are the alternatives equal? $\endgroup$ – Dan Brumleve Jan 24 '12 at 11:02
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    $\begingroup$ @Gigili, no, -1 and 6. $\endgroup$ – lhf Jan 24 '12 at 11:12
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Consider any point $x\in\mathbb{R}$ and assume that $f$ is continuous at $x$. You can find two sequences $\{a_n\}\subset\mathbb{Q}$ and $\{b_n\}\subset\mathbb{R}\setminus\mathbb{Q}$ such that $\lim a_n=\lim b_n=x$ (do you know why they exist and/or how to find those?). Now use the Heine property for continuity to say that $$5x=\lim 5a_n=\lim f(a_n)=f(\lim a_n)=f(\lim b_n)=\lim f(b_n)=\lim b_n^2-6=x^2-6$$ Now you can find $x$.

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    $\begingroup$ And having found $x$ you must still prove that the function is continuous at $x$, since using just two sequences won't prove it. $\endgroup$ – Marc van Leeuwen Jan 24 '12 at 11:18

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