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)(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))#
We want: #abs((x/4+3)-9/2) < epsilon#
Look at the thing we want to make small. Rewrite this, looking for the thing we control.
#abs((x/4+3)-9/2) =abs(x/4+6/2-9/2)#
# = abs(x/4-3/2) #
# = abs((x-6)/4)#
# = (abs(x-6))/4#
In order to make this less than #epsilon#, it suffices to make #abs(x-6)# less than #4epsi#.
Note that:
If #abs(x-6) < 4 epsilon#, then we can divide both sides by #4#. (Since #4# is positive, we don't need to reverse the inequalities.) to get #abs(x-6)/4 < epsilon#
Proving our L is correct -- Writing the proof
Claim: #lim_(xrarr6)(x/4+3) = 9/2#
Proof:
Given #epsilon > 0#, choose #delta = 4epsilon#. (Note that #delta# is positive.)
Now if #0 < abs(x-6) < delta# then
#abs((x/4+3)-9/2) = abs(x/4-3/2)#
# = abs((x-6)/4)#
# = abs(x-6)/4#
# < delta/4#
# = (4epsilon)/4#
# = epsilon#
We have shown that for any positive #epsilon#, there is a positive #delta# such that for all #x#, if #0 < abs(x-6) < delta#, then #abs((x/4+3)-9/2) < epsilon#.
So, by the definition of limit, we have #lim_(xrarr6)(x/4+3) = 9/2#.
