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Math help
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
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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Suuper W, SuuperPR2
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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  • Ali
idk lol why r u asking me
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