Current Events > Who is ready for a brutal calculus (real analysis) limit problem?

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FLOUR
03/18/21 5:22:36 PM
#1:


Note this is just a warm up. Define f(x) as follows. f(x)=0 if x is irrational and f(x)=1 if x is rational. Prove that the limit does not exist at any point. Reminder this is just the warm up problem. It gets worse.

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nothanks1
03/18/21 5:24:02 PM
#2:


I could ask one of the math teachers
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UnholyMudcrab
03/18/21 5:24:56 PM
#3:


Something something differential, something something x approaches y
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Phynaster
03/18/21 5:25:44 PM
#4:


f(x) is the room outside

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Rebel
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SPE
03/18/21 5:25:53 PM
#5:


x = 0/f

x = 0/f

rational and irrational merge and form a ying yang, so x = 0f0

0f0 means sun shadows smiley

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#6
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NightRender
03/18/21 5:30:26 PM
#7:


It's something like, for whatever c x is approaching, there is some number within episilon of it that is the opposite rationality of f(c), so can't shrink delta more than 1.

That can probably be cleaned up and worded smarterly. I haven't written a proper proof in years, let alone an epsilon delta proof.

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