Finding the values to use in 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)(1/3))color(red)((1-9x^2)/(1-3x) = color(blue)(2)#
So we want to make #abs(underbrace(color(red)((1-9x^2)/(1-3x) ))_(color(red)(f(x)) )-underbrace(color(blue)(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)((1/3)))_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((1-9x^2)/(1-3x) - 2) = abs(((1-3x)(1+3x))/(1-3x) - 2)#
# = abs(((1+3x) - 2)#
#=abs(3x-1)#
Recall that we control the size of #abs(x-(1/3))# and if we factor a positive #3# out of the last expression we get
#=3abs(x-1/3)#
In order to make this less than #epsilon#, it suffices to make #abs(x-(1/3))# less than #epsi/3#
Writing the proof
Claim: #lim_(xrarr1/3)((1-9x^2)/(1-3x) ) = 2#
Proof:
Given #epsilon > 0#, choose #delta = epsilon/3#. (Note that #delta# is positive.)
Now if #0 < |x-(1/3)| < delta# then
#abs((1-9x^2)/(1-3x) -2) = abs(((1-3x)(1+3x))/(1-3x) - 2)#
# = abs(((1+3x) - 2)#
#=abs(3x-1)#
# = abs(3)abs(x-1/3)#
# = 3 abs(x-1/3)#
# < 3 delta#
# = 3 (epsi/3)#
# = epsilon#
We have shown that for any positive #epsilon#, there is a positive #delta# such that for all #x#, if #0 < abs(x-(1/3)) < delta#, then #abs((1-9x^2)/(1-3x) - 2) < epsilon#.
So, by the definition of limit, we have #lim_(xrarr1/3)((1-9x^2)/(1-3x) ) = 2#.
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-1/3))#, then a choice of #x = 1/3# would result in an undefined expression. #abs((1-9x^2)/(1-3x) - 2) #.
