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)((3x+5)))_(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)) which is equal to abs(x+2).
Look at the thing we want to make small. We want to see the thing we control in the thing we want small..
abs((3x+5)-(-1)) = abs(3x+6) = abs (3(x+2)) =abs(3)abs(x+2) =3abs(x+2)
We want to make 3abs(x+2) small and we control the size of
abs(x+2).
We can make 3abs(x+2) < epsilon by making abs(x+2) < epsilon/3.
So we will choose delta = epsilon/3. (Any lesser delta would also work.)
(Detail: if abs(x+2) < epsilon/3, then we can multiply on both sides by the positive number 3 to get 3abs(x+2) < epsilon.)
Now we need to actually write up the proof:
Proof
Given epsilon > 0, choose delta = epsilon/3. " " (note that delta is also positive).
Now for every x with 0 < abs(x-(-2)) < delta, we have
abs(x + 2) < delta, and
abs((3x+5) - (-1)) = abs(3x+6)) = abs(3)abs(x+2) = 3abs(x+2) < 3delta
[Detail: if abs(x+2) < delta, we can conclude that 3abs(x+2) < 2delta. " " We usually do not mention this, but leave it to the reader. See below.]
And 3 delta = 3 epsilon/3 = epsilon
Therefore, with this choice of delta, whenever 0 < abs(x-(-2)) < delta, we have abs((3x+5) - (-1)) < epsilon
So, by the definition of limit, lim_(xrarr-2)(3x+5) = -1.
We can condense a bit
Given epsilon > 0, choose delta = epsilon/3. " " (note that delta is also positive).
for every x with 0 < abs(x-(-2)) < delta, we have
abs((3x+5)-1) = abs(3x+6)
= abs(3(x-2))
= 3abs(x-2)
< 3delta = 3 epsilon/3 = epsilon.
So, abs((3x+5)-(-1)) < epsilon.
