anonymous
  • anonymous
How to do epsilon-delta proof of lim_(x->0) (x^2-2x+3)=3?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
you want to show that given any ϵ>0 you can find δ so that if |x−3|<δ you get |x2−2x+3−3|<ϵ as usual we work backwards
anonymous
  • anonymous
|x−0|<δ?
anonymous
  • anonymous
\[|x^2-2x+3-3|=|x^2-2x|=|x||x-2|\] and we get to control |x|

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anonymous
  • anonymous
yes i was being stupid i know, but i just wanted it to look like \[|x-a|<\delta\] in this case of course \(a=0\)
anonymous
  • anonymous
now since we control |x| by making it as small as we like, we only have to get a handle on |x-2|
anonymous
  • anonymous
Yep.
anonymous
  • anonymous
so the usual trick is since x is near zero we can say assume that |x|<1 so |x-2|<3
anonymous
  • anonymous
|x-2| < C
anonymous
  • anonymous
|x-2| < 1 and -1 < x < 1 -3 < x-2 < -1
anonymous
  • anonymous
this line was wrong i meant \[|x||x-2|< \frac{\epsilon}{3}\times 3=\epsilon\]
anonymous
  • anonymous
satellite are u sure that |x - 2|<3 ? Cuz, if |X|<1 so |x-2|< 1-2 -> |x-2|< -1
anonymous
  • anonymous
let me write it again correctly assume |x|<1 so -1
anonymous
  • anonymous
we want an upper bound for the term |x-2| and i got 3 you can of course find a different one but now that i have bounded |x-2|<3 i can pick \(\delta=\frac{\epsilon}{3}\)
anonymous
  • anonymous
therefore is \(|x|<\frac{\epsilon}{3}\) and \(\frac{\epsilon}{3}<1\) we get \[|x||x-2|<\frac{\epsilon}{3}\times 3=\epsilon\]
anonymous
  • anonymous
satellite, is it honest?
anonymous
  • anonymous
?
anonymous
  • anonymous
i'm talking about mathematical honesty in this part: assume |x|<1 so -1

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