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)((x/4+3)))_(color(red)(f(x)) )-underbrace(color(blue)(9/2))_color(blue)(L))# less than some given #epsilon# and we control (through our control of #delta#) the size of #abs(x-underbrace(color(green)(6))_color(green)(a))#
Look at the thing we want to make small:
#abs((x/4+3)-9/2) = abs (x/4-3/2) = abs((x-6)/4) = abs(x-6)/abs4 = abs(x-6)/4#
And there's the thing we control, in the numerator!
We can make #abs(x-6)/4 < epsilon# by making #abs(x-6) < 4epsilon#.
So we will choose #delta = 4 epsilon#. (Any lesser #delta# would also work.)
(Detail: if #abs(x-6) < 4epsilon#, then we can multiply on both sides by the positive number #1/4# to get #abs(x-6)/4 < epsilon#.)
Now we need to actually write up the proof:
Proof
Given #epsilon > 0#, choose #delta = 4epsilon#. #" "# (note that #delta# is also positive).
Now for every #x# with #0 < abs(x-6) < delta#, we have
#abs(f(x)-9/2) = abs((x/4+3)-9/2) = abs((x-6)/4) = abs(x-6)/4 < delta/4#
[Detail if #abs(x-6) < delta#, we can conclude that #abs(x-6)/4 < delta/4#. -- we usually do not mention this, but leave it to the reader. See below.]
And #delta /4 = (4epsilon)/4 = epsilon#
Therefore, with this choice of delta, whenever #0 < abs(x-6) < delta#, we have #abs(f(x)-9/2) < epsilon#
So, by the definition of limit, #lim_(xrarr6)(x/4+3) = 9/2#.
We can condense a bit
for every #x# with #0 < abs(x-6) < delta#, we have
#abs(f(x)-9/2) = abs((x/4+3)-9/2)#
# = abs((x-6)/4)#
# = abs(x-6)/4#
# < delta/4 = (4epsilon)/4 = epsilon#.
So, #abs(f(x)-9/2) < epsilon#.
