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enter image description hereI picked out the points $-1,0,1,2,3$

Would the end points $-1$ and $3$ be also considered as points of discontinuity? The answer says $-1$ isn’t but $3$ is

Aditya
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1 Answers1

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Since $|1-x|$ is continuous everywhere, $f$ is discontinuous in every point where $x \mapsto [x]$ is.

So $-1$ is not a point of discontinuity of $f$ because $x \mapsto [x]$ is continuous from the right in $-1$. But $3$ is a point of discontinuity of $f$ because $x \mapsto [x]$ is not continuous from the left in $3$.

TheSilverDoe
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  • I checked the graph, and it breaks at $x=-1$. Also the RHL and the LHL dont match – Aditya Aug 16 '20 at 11:43
  • It breaks, but since the function is defined on $[-1, 3]$, you just have to check what happens at the right of $-1$. The left side doesn't interest you because it is not in the domain of $f$. – TheSilverDoe Aug 16 '20 at 11:44
  • Yes but the function breaks at $x=-1$, so that should count right? – Aditya Aug 16 '20 at 11:50
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    No. It breaks only if you consider it on an interval that contains $-1$ in its interior. Since $-1$ is the left point of the domain on $f$, you just have to check if the limit of $f$ when $x$ tends to $-1$ from the right is equal to $f(-1)$. – TheSilverDoe Aug 16 '20 at 11:53
  • For example, the function $x \mapsto [x]$ is continuous on $[-1, 0)$. But it is not continuous on $(-2,0)$ because it is discontinuous at $x=-1$. – TheSilverDoe Aug 16 '20 at 11:55
  • Okay but in that case the function is continuous to the left of 3 as well – Aditya Aug 16 '20 at 12:05
  • No. For the function $g : x \mapsto [x]$, you have $g(3)=3$ but $g(3-\varepsilon)=2$ for all $\varepsilon > 0$. So $g$ is not continuous from the left at $x=3$. However, $g(-1)=-1$ and $g(-1+\varepsilon)=-1$ for all $\varepsilon$, so $g$ is continuous from the right at $x=-1$. – TheSilverDoe Aug 16 '20 at 12:08
  • I uploaded the graph. It is continuous to the left of 3 – Aditya Aug 16 '20 at 12:22
  • I meant $g(-1+\varepsilon)=-1$ for all $\varepsilon \in (0,1)$. And see the graph, the function is not continuous at $x=3$ : indeed the limit from the left is not equal to the value of the function at the point. – TheSilverDoe Aug 16 '20 at 13:28
  • I get it now, thanks – Aditya Aug 16 '20 at 15:35