Find the limit
When #x# is close to #-2#, then #x-2# is close to #-4#, so
#abs(x-2)# is close to #abs(-4) = 4#
So I claim that #lim_(xrarr-2)abs(x-2) = 4#
Proving our L is correct -- Finding the proof
This explanation of finding the proof is a bit long. If you just want to read the proof, scroll down.
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#.
So we want to make #abs(underbrace(color(red)(abs(x-2)))_(color(red)(f(x)) )-underbrace(color(blue)(4))_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 need to understand the absolute value function. Note that
#abs(x-2) = {(x-2,"if",x-2 >= 0),(-(x-2),"if",x-2 < 0) :}#.
We are interested in what happens when #x# is close to #-2# where we will have #abs(x-2) = -(x-2)#.
Let's make sure that when #abs(x-(-2)) < delta#, we will have #x < 2#.
Notice that #abs(x-(-2)) = abs(x+2)#.
To get #x - 2 < 0# we simply need #x +2 < 4# , so we'll make sure that #delta <= 4#.
We can now rewrite what we want.
We want: #abs(-(x-2)-4) < epsilon# #" "# (for #x < 2#)
But #abs(-(x-2)-4) =abs(-x-2) = abs(-1)abs(x+2) = abs(x+2)# when #abs(x+2) < delta#.
So we need only make #delta <= epsilon# to make the proof work.
Proving our L is correct -- Writing the proof
Claim: #lim_(xrarr-2)abs(x-2) = 4#
Proof:
Given #epsilon > 0#, choose #delta = min{4, epsilon}#. (Note that #delta# is positive.)
Now if #0 < abs(x-(-2)) < delta# then #x+2 < 4#, so #x-2 < 0# and #abs(x-2) = -(x-2)#.
Therefore, if #0 < abs(x-(-2)) < delta#, then
#abs(abs(x-2)-4) = abs(-(x-2)-4) =abs(-x-2) = abs(-1)abs(x+2) = abs(x+2)# which is less than #delta# which is less than or equal to #epsilon#.
Finally, then, if #0 < abs(x-(-2)) < delta#, then #abs(abs(x-2)-4) < 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(abs(x-2)-4) < epsilon#.
So, by the definition of limit, we have #lim_(xrarr-2)abs(x-2) = 4#
