Question

How do you prove that the limit of `(x^2 - 4x + 5) = 1` as x approaches 2 using the epsilon delta proof?

Answer 1

See below.

Explanation:

Problem:
`lim_(x->2)(x^2-4x+5)=1`

Work:
`|(x^2-4x+5)-1|< epsilon`
`|x^2-4x+4|< epsilon`
`|(x-2)|^2< epsilon`
`|x-2|< sqrt(epsilon)`

Proof:

`AA` `epsilon>0`, `EE` `delta>0` such that
if `0<|x-2|< delta`, then `|(x^2-4x+5)-1|< epsilon`
Given `0<|x-2|< delta`, let `delta=sqrt(epsilon)`
`|x-2|< sqrt(epsilon)`
`|x-2|^2< epsilon`
`|(x-2)^2|< epsilon`
`|x^2-4x+4|< epsilon`
`|(x^2-4x+5)-1|< epsilon`
`thereforelim_(x->2)(x^2-4x+5)=1`