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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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
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anonymous
  • anonymous
?
anonymous
  • anonymous
sorry. that last response was supposed to go to somebody else, but for some reason got put here. sorry about that!

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anonymous
  • anonymous
uaheuhuheuaehhhaeuhuehuhe
anonymous
  • anonymous
i gave you answer once already
experimentX
  • experimentX
at x's tends to 0, all terms of function tends to 0 .... except constant 3, so 3
anonymous
  • anonymous
No, you did not.
anonymous
  • anonymous
i did
anonymous
  • anonymous
I know that: 0<|x - 0| < delta, so |f(x) -3|< epsilon
anonymous
  • anonymous
|x^2 -2x +3 - 3| < epsilon
anonymous
  • anonymous
|x^2 - 2x| < epsilon |x(x-2)| = |x| |x-2| = epsilon C Taking |x-2| < C epsilon/C = delta And now what i do?
experimentX
  • experimentX
|x| |x-2| < e => |x - 0| < e / |x-2| < d (suppose it) => 3 is limit as x->0
anonymous
  • anonymous
Sorry, I did not understand.
experimentX
  • experimentX
beginning with |x^2 -2x +3 - 3| < epsilon show |x - 0| < d
anonymous
  • anonymous
I need a answer with more formalization.

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