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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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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 --- Welcome to the Spore Store! My shrooms are delicious, suspicious, but never malicious! ... Copied to Clipboard!
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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. --- Dedicated to D - 4/15/05 ... Copied to Clipboard!
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