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2018-12-23 05:57:09 (This post was last modified: 2018-12-23 05:57:58 by Ilraon.) Hi there, long time no see huh?

I'm seeking for help concerning this question:

is the function defined for all t in ]0,1] by f(t)=t*floor(1/t) piecewise continuous?

My teacher integrated it in a lesson about integration, and we have only seen the integration of piecewise continuous functions.

Hope you're all doing well

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Why are you dealing with functions that contain rounding (or truncating) functions? How do you integrate that?

Anyway I'd think it's not continuous, as there'd be a gap where the output of the floor function changes.

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Pick any point x in [0, 1]. Then we have two possibilities:

1) 1/x is not an integer. In that case, the function is continuous at x.

2) 1/x is an integer. In that case, the function makes a jump at x. But you know that the function on both sides must be continuous, since integers are discrete — the values 1/(1/x + 0.5) and 1/(1/x - 0.5) lie on either side of x, and the variable doesn't take any inverse integer values in that range other than x (in other words, for all h in [1/(1/x + 0.5), 1/(1/x - 0.5)] so that h ≠ x, 1/h is not an integer), meaning that the function is continuous in a deleted neighbourhood around x.

That way, for all x, you can show that the function is continuous in a deleted neighbourhood around x. That makes the function piecewise continuous.

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idk lol why r u asking me

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