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megafaggot !LC028coGd. at Wed, 9 Oct 2024 01:12:06 UTC No. 16416662
hiiii im in calc 3 nd im trying to practice Epsilon-Delta proofs. can som1 pls help me nd look at my work idk if it's right. my main questions r:
i know that converting into polar doesn't prove the limit of the original function but since it's not dependent on theta would this proof work?
since r can't b negative i added a "from above" to the limit but i wasn't sure how it'd affect the proof. is there anything that'd change that i missed?
finally, im jus trying to understand, so delta is any function of epsilon? is what i chose for delta correct?
ty for any help !!
megafaggot !LC028coGd. at Wed, 9 Oct 2024 01:14:10 UTC No. 16416667
>>16416662
oh also under suppose it should b |r|, not |x| :p
Anonymous at Wed, 9 Oct 2024 01:16:27 UTC No. 16416671
>>16416662
>Epsilon-Delta proofs
not math
Anonymous at Wed, 9 Oct 2024 01:21:41 UTC No. 16416677
>>16416662
Polar substitution is fine here
megafaggot !LC028coGd. at Wed, 9 Oct 2024 01:49:09 UTC No. 16416708
>>16416671
huh why not sry i jus assumed i could ask this here
>>16416677
bc it's only dependent on r nd not theta?
Anonymous at Wed, 9 Oct 2024 02:20:52 UTC No. 16416744
>>16416662
looks decent anon
>doesn’t prove the limit of the original function
in this case it does, but if you want an extra step to verify this, insert
>suppose 0<|x|<delta and 0<|y|<delta
>then 0<x^2+y^2<delta^2
and changing the definition of delta to compensate
Anonymous at Wed, 9 Oct 2024 02:44:01 UTC No. 16416772
>>16416671
not a real woman
Anonymous at Wed, 9 Oct 2024 03:03:07 UTC No. 16416780
>>16416662
Are you a girl?
megafaggot !LC028coGd. at Wed, 9 Oct 2024 03:47:55 UTC No. 16416848
>>16416780
im a trans girl, why
Anonymous at Wed, 9 Oct 2024 04:46:05 UTC No. 16416939
>>16416708
>bc it's only dependent on r nd not theta?
Because you're evaluating the limit at (0,0), and "(x,y) -> (0,0)" is the same as "|(x,y)| -> 0"
Anonymous at Wed, 9 Oct 2024 05:33:20 UTC No. 16416972
>tripping on a homework question
Anyways, logarithms blow up when your input is less than 1, so this proof is incorrect. I would split up the cases for epsilon greater than or equal to 1 and epsilon less than 1.
megafaggot !LC028coGd. at Wed, 9 Oct 2024 06:07:37 UTC No. 16417004
>>16416972
yeahh thats what i was jus looking at. i think i messed up by doing |lnr| < |x| b/c x > lnx because thats not true for 0<x<1
megafaggot !LC028coGd. at Wed, 9 Oct 2024 14:11:34 UTC No. 16417498
>>16416744
i added sqrt(x^2+y^2)=r to the definition but how would this change delta? if i substituted in sqrt(x^2+y^2) for every r, wouldn't it b the exact same?
>>16416972
i think i fixed it by bounding delta, im unsure of the (1/r) > -ln(r) step but i think it works if 0<r<1. also this isn't a hw problem i was jus doing it for fun lol
Anonymous at Wed, 9 Oct 2024 14:12:46 UTC No. 16417501
>homework thread
Anonymous at Wed, 9 Oct 2024 14:13:39 UTC No. 16417505
>>16416662
>Episolon Delta proofs
>In calc 3
What the actual fuck is wrong with the math department
megafaggot !LC028coGd. at Wed, 9 Oct 2024 14:48:14 UTC No. 16417551
>>16417505
>also this isn't a hw problem i was jus doing it for fun