**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#.