The explanation has two sections. There is a preliminary analysis to find the values used in the proof, then there is a presentation of the proof itself.
Finding the proof
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#.
We have been asked to show that
#lim_(xrarrcolor(green)(-1.5))color(red)((9-4x^2)/(3+2x) = color(blue)(6)#
So we want to make #abs(underbrace(color(red)((9-4x^2)/(3+2x) ))_(color(red)(f(x)) )-underbrace(color(blue)(6))_color(blue)(L))# less than some given #epsilon# and we control (through our control of #delta#) the size of #abs(x-underbrace(color(green)((-1.5)))_color(green)(a))#
We want: #abs((9-4x^2)/(3+2x) - 6) < epsilon#
Look at the thing we want to make small. Rewrite this, looking for the thing we control.
#abs((9-4x^2)/(3+2x) - 6) = abs(((3-2x)(3+2x))/(3+2x) - 6)#
# = abs(((3-2x) - 6)#
#=abs(-2x-3)#
Recall that we control the size of #abs(x-(-1.5)) = abs(x-(-3/2))#
I see #abs(x+3# which is the same as #abs(x-(-3))# so let's 'factor out #-2#.
# = abs(-2)abs(x+3/2)#
# = 2 abs(x-(-3/2))#
In order to make this less than #epsilon#, it suffices to make #abs(x-(-1.5))# less than #epsi/2#
Proving our L is correct -- Writing the proof
Claim: #lim_(xrarr-1.5)((9-4x^2)/(3+2x) ) = 6#
Proof:
Given #epsilon > 0#, choose #delta = epsilon/2#. (Note that #delta# is positive.)
Now if #0 < |x-(-1.5)|=|x-(-3/2)|< delta# then
#abs((9-4x^2)/(3+2x) -6) = abs(((3-2x)(3+2x))/(3+2x) - 6)#
# = abs(((3-2x) - 6)#
#=abs(-2x-3)#
# = abs(-2)abs(x+3/2)#
# = 2 abs(x-(-3/2))#
# < 2 delta#
# = 2 (epsi/2)#
# = epsilon#
We have shown that for any positive #epsilon#, there is a positive #delta# such that for all #x#, if #0 < abs(x-(-1.5)) < delta#, then #abs((9-4x^2)/(3+2x) -6) < epsilon#.
So, by the definition of limit, we have #lim_(xrarr-1.5)((9-4x^2)/(3+2x) ) = 6#.
Note
This is an example of a limit in which the strict inequality #0 < abs(x-delta)# is very important. If we allowed #0 = abs(x-(-1.5))#, then a choice of #x = -1.5# would result in an undefined expression. #abs((9-4x^2)/(3+2x) -6) #.
