Question

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

Answer 1

http://www.classicshorts.com/stories/cwrdmaup.html

Answer 2

?

Answer 3

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

Answer 4

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Answer 5

i gave you answer once already

Answer 6

at x's tends to 0, all terms of function tends to 0 .... except constant 3, so 3

Answer 7

No, you did not.

Answer 8

i did

Answer 9

I know that:

0<|x - 0| < delta, so |f(x) -3|< epsilon

Answer 10

|x^2 -2x +3 - 3| < epsilon

Answer 11

|x^2 - 2x| < epsilon

|x(x-2)| = |x| |x-2| = epsilon C

Taking |x-2| < C

epsilon/C = delta

And now what i do?

Answer 12

|x| |x-2| < e => |x - 0| < e / |x-2| < d (suppose it) => 3 is limit as x->0

Answer 13

Sorry, I did not understand.

Answer 14

beginning with |x^2 -2x +3 - 3| < epsilon
show |x - 0| < d

Answer 15

I need a answer with more formalization.