How do you prove the statement lim as x approaches 2 for `(x^2 - 4x + 5) = 1` using the epsilon and delta definition?

The explanation has two sections. There is a preliminary analysis to find the values used in the proof, then there is a presentation of the proof itself.

Finding the proof

By definition,

`lim_(xrarrcolor(green)(a))color(red)(f(x)) = color(blue)(L)` if and only if

for every `epsilon > 0`, there is a `delta > 0` such that:
for all `x`, `" "` if `0 < abs(x-color(green)(a)) < delta`, then `abs(color(red)(f(x))-color(blue)(L)) < epsilon`.

We have been asked to show that

`lim_(xrarrcolor(green)(2))color(red)(x^2-4x+5) = color(blue)(1)`

So we want to make `abs(underbrace(color(red)(x^2-4x+5))_(color(red)(f(x)) )-underbrace(color(blue)(1))_color(blue)(L))` less than some given `epsilon` and we control (through our control of `delta`) the size of `abs(x-underbrace(color(green)((2)))_color(green)(a))`

We want: `abs((x^2-4x+5)-1) < epsilon`

Look at the thing we want to make small. Rewrite this, looking for the thing we control.

`abs((x^2-4x+5)-1) =abs(x^2-4x+4)`

`= abs((x-2)^2)`

`= (x-2)^2`

In order to make this less than `epsilon`, it suffices to make `abs(x-2)` less than `sqrtepsi`

Proving our L is correct -- Writing the proof

Claim: `lim_(xrarr2)(x^2-4x+5) = 1`

Proof:

Given `epsilon > 0`, choose `delta = sqrtepsilon`. (Note that `delta` is positive.)

Now if `0 < abs(x-2) < delta` then

`abs((x^2-4x+5)-1) =abs(x^2-4x+4)`

`= abs((x-2)^2)`

`= (x-2)^2`

`< delta^2` `" "` (See Note below)

`= (sqrtepsilon)^2`

`= epsilon`

We have shown that for any positive `epsilon`, there is a positive `delta` such that for all `x`, if `0 < abs(x-2) < delta`, then `abs((x^2-4x+5)-1) < epsilon`.

So, by the definition of limit, we have `lim_(xrarr2)(x^2-4x+5) = 1`.

Note

Since the squaring function is increasing on positive values,

`abs(x-2) < sqrtdelta` implies that `abs(x-2)^2 = (x-2)^2 < delta^2`