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.
