Preliminary analysis
#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)((5-2x)))_(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))#
Look at the thing we want to make small. We want to see the thing we control.
#abs((5-2x)-1) = abs(-2x+4) = abs (-2(x-2)) =abs(-2)abs(x-2) =2abs(x-2)#
And there's #abs(x-2)#, the thing we control
We can make #2abs(x-2) < epsilon# by making #abs(x-2) < epsilon/2#.
So we will choose #delta = epsilon/2#. (Any lesser #delta# would also work.)
(Detail: if #abs(x-2) < epsilon/2#, then we can multiply on both sides by the positive number #2# to get #2abs(x-2) < epsilon#.)
Now we need to actually write up the proof:
Proof
Given #epsilon > 0#, choose #delta = epsilon/2#. #" "# (note that #delta# is also positive).
Now for every #x# with #0 < abs(x-2) < delta#, we have
#abs(f(x)-1) = abs((5-2x) - 1) = abs(-2x+4)) = abs(-2)abs(x-2) = 2abs(x-2) < 2delta#
[Detail if #abs(x-2) < delta#, we can conclude that #2abs(x-2) < 2delta#. #" "# We usually do not mention this, but leave it to the reader. See below.]
And #2 delta = 2 epsilon/2 = epsilon#
Therefore, with this choice of delta, whenever #0 < abs(x-2) < delta#, we have #abs(f(x) - 1) < epsilon#
So, by the definition of limit, #lim_(xrarr2)(5-2x) = 1#.
We can condense a bit
for every #x# with #0 < abs(x-6) < delta#, we have
#abs(f(x) - 1) = abs((5-2x)-1)#
# = abs(-2(x-2))#
# = 2abs(x-2)#
# < 2delta = 2 epsilon/2 = epsilon#.
So, #abs(f(x) - 1) < epsilon#.
