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#