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

Question

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

anonymous · Answer

anonymous · Answer

?

anonymous · Answer

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

anonymous · Answer

uaheuhuheuaehhhaeuhuehuhe

anonymous · Answer

i gave you answer once already

experimentx · Answer

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

anonymous · Answer

No, you did not.

anonymous · Answer

i did

anonymous · Answer

I know that:

0<|x - 0| < delta, so |f(x) -3|< epsilon

anonymous · Answer

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

anonymous · Answer

|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 · Answer

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

anonymous · Answer

Sorry, I did not understand.

experimentx · Answer

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

anonymous · Answer

I need a answer with more formalization.