Question

.

Answer 1

the correct funciton is:
(x^2-2x+3)

Answer 2

you probly suspect that the limit is -3, so
for x-0<delta (x^2-2x-3 -(-3))<epsilon
which is true

Answer 3

in this case x is you delta

Answer 4

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

Answer 5

do the same:
you probly suspect that the limit is 3, so
for x-0=x<delta (x^2-2x+3 -3))=x^2-2x <epsilon
which is true

Answer 6

Yes I did it.

Answer 7

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