Question

How to do epsilon-delta proof of lim_(x->0) (x^2-2x+3)=3?

Answer 1

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

Answer 2

|x−0|<δ?

Answer 3

\[|x^2-2x+3-3|=|x^2-2x|=|x||x-2|\] and we get to control |x|

Answer 4

yes i was being stupid i know, but i just wanted it to look like
\[|x-a|<\delta\] in this case of course \(a=0\)

Answer 5

now since we control |x| by making it as small as we like, we only have to get a handle on |x-2|

Answer 6

Yep.

Answer 7

so the usual trick is since x is near zero we can say assume that |x|<1 so |x-2|<3

Answer 8

|x-2| < C

Answer 9

|x-2| < 1 and

-1 < x < 1

-3 < x-2 < -1

Answer 10

this line was wrong i meant
\[|x||x-2|< \frac{\epsilon}{3}\times 3=\epsilon\]

Answer 11

satellite are u sure that |x - 2|<3 ? Cuz, if |X|<1 so |x-2|< 1-2 -> |x-2|< -1

Answer 12

let me write it again correctly
assume |x|<1 so -1<x<1 and therfore -3<x-2<-1 ensuring |x-2|<3

Answer 13

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 \(\delta=\frac{\epsilon}{3}\)

Answer 14

therefore is \(|x|<\frac{\epsilon}{3}\) and \(\frac{\epsilon}{3}<1\) we get
\[|x||x-2|<\frac{\epsilon}{3}\times 3=\epsilon\]

Answer 15

satellite, is it honest?

Answer 16

?

Answer 17

i'm talking about mathematical honesty in this part: assume |x|<1 so -1<x<1 and therfore -3<x-2<-1 ensuring |x-2|<3
I'm confused