Preliminary analysis for the proof
We want to show that #lim_(xrarr-4)(1/2x-1) = -3#.
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)((1/2x-1)))_(color(red)(f(x)) )-underbrace(color(blue)((-3)))_color(blue)(L))# less than some given #epsilon# and we control (through our control of #delta#) the size of #abs(x-underbrace(color(green)((-4)))_color(green)(a)) = abs(x+4)#
Look at the thing we want to make small. Rewrite this, looking for the thing we control.
#abs((1/2x-1)-(-3)) = abs(1/2x + 2)#
We want to see the thing we control, which is #abs(x+4)#. Let's make it appear by factoring #1/2#.
# = abs(1/2(x+4))= abs(1/2)abs(x+4) = 1/2abs(x+4) #
We can make #1/2abs(x+4) < epsilon# by making #abs(x+4) < 2 epsilon#..
So we will choose #delta = 2epsilon#. (Any lesser #delta# would also work.)
Now we need to actually write up the proof:
Proof
Given #epsilon > 0#, choose #delta = 2epsilon#. #" "# (note that #delta# is also positive).
Now for every #x# with #0 < abs(x-(-4)) < delta#, we have
#abs (x+4) < delta# So,
#abs((1/2x-1)-(-3)) = abs(1/2x+2) = 1/2 abs(x+4) < 1/2delta = 1/2(2epsilon) = epsilon#
Therefore, with this choice of delta, whenever #0 < abs(x-(-4)) < delta#, we have #abs((1/2x-1) - (-3)) < epsilon#
So, by the definition of limit, #lim_(xrarr-1)(1/2x-1) = -3#.
